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인기 있는 삼각법 >

tan^2(x)= 1/(cos(x)+1)

해법

tan^{2} (x) = \frac{1}{cos ( x ) + 1}

해법

x = 0.64026 \dots + 2 πn, x = 2 π - 0.64026 \dots + 2 πn, x = 2.15910 \dots + 2 πn, x = - 2.15910 \dots + 2 πn

+1

도

x = 36.68445 \dots^{\circ} + 36 0^{\circ} n, x = 323.31554 \dots^{\circ} + 36 0^{\circ} n, x = 123.70783 \dots^{\circ} + 36 0^{\circ} n, x = - 123.70783 \dots^{\circ} + 36 0^{\circ} n

솔루션 단계

tan^{2} (x) = \frac{1}{cos ( x ) + 1}

양쪽을 제곱

(tan^{2} (x))^{2} = (\frac{1}{cos ( x ) + 1})^{2}

빼다

(\frac{1}{cos ( x ) + 1})^{2}

양쪽에서

tan^{4} (x) - \frac{1}{( cos ( x ) + 1 ) ^{2}} = 0

tan^{4} (x) - \frac{1}{( cos ( x ) + 1 ) ^{2}}

단순화하세요:

\frac{tan ^{4} ( x ) ( cos ( x ) + 1 ) ^{2} - 1}{( cos ( x ) + 1 ) ^{2}}

tan^{4} (x) - \frac{1}{( cos ( x ) + 1 ) ^{2}}

요소를 분수로 변환:

tan^{4} (x) = \frac{tan ^{4} ( x ) ( cos ( x ) + 1 ) ^{2}}{( cos ( x ) + 1 ) ^{2}}

= \frac{tan ^{4} ( x ) ( cos ( x ) + 1 ) ^{2}}{( cos ( x ) + 1 ) ^{2}} - \frac{1}{( cos ( x ) + 1 ) ^{2}}

분모가 같기 때문에, 분수를 합친다:

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{tan ^{4} ( x ) ( cos ( x ) + 1 ) ^{2} - 1}{( cos ( x ) + 1 ) ^{2}}

\frac{tan ^{4} ( x ) ( cos ( x ) + 1 ) ^{2} - 1}{( cos ( x ) + 1 ) ^{2}} = 0

\frac{f ( x )}{g ( x )} = 0 \Rightarrow f (x) = 0

tan^{4} (x) (cos (x) + 1)^{2} - 1 = 0

삼각성을 사용하여 다시 쓰기

- 1 + (1 + cos (x))^{2} tan^{4} (x)

기본 삼각형 항등식 사용:

cos (x) = \frac{1}{sec ( x )}

= - 1 + (1 + \frac{1}{sec ( x )})^{2} tan^{4} (x)

- 1 + (1 + \frac{1}{sec ( x )})^{2} tan^{4} (x) = 0

- 1 + (1 + \frac{1}{sec ( x )})^{2} tan^{4} (x)

요인:

(tan^{2} (x) (1 + \frac{1}{sec ( x )}) + 1) (tan^{2} (x) (1 + \frac{1}{sec ( x )}) - 1)

- 1 + (1 + \frac{1}{sec ( x )})^{2} tan^{4} (x)

- 1 + (1 + \frac{1}{sec ( x )})^{2} tan^{4} (x) - 1 + ((1 + \frac{1}{sec ( x )}) tan^{2} (x))^{2}

로 다시 씁니다

- 1 + (1 + \frac{1}{sec ( x )})^{2} tan^{4} (x)

지수 규칙 적용:

a^{b c} = (a^{b})^{c}

tan^{4} (x) = (tan^{2} (x))^{2}

= - 1 + (1 + \frac{1}{sec ( x )})^{2} (tan^{2} (x))^{2}

지수 규칙 적용:

a^{m} b^{m} = (ab)^{m}

(1 + \frac{1}{sec ( x )})^{2} (tan^{2} (x))^{2} = ((1 + \frac{1}{sec ( x )}) tan^{2} (x))^{2}

= - 1 + ((1 + \frac{1}{sec ( x )}) tan^{2} (x))^{2}

= - 1 + ((1 + \frac{1}{sec ( x )}) tan^{2} (x))^{2}

두 제곱 공식의 차이 적용:

x^{2} - y^{2} = (x + y) (x - y)

- 1 + ((1 + \frac{1}{sec ( x )}) tan^{2} (x))^{2} = ((1 + \frac{1}{sec ( x )}) tan^{2} (x) + 1) ((1 + \frac{1}{sec ( x )}) tan^{2} (x) - 1)

= ((1 + \frac{1}{sec ( x )}) tan^{2} (x) + 1) ((1 + \frac{1}{sec ( x )}) tan^{2} (x) - 1)

(tan^{2} (x) (1 + \frac{1}{sec ( x )}) + 1) (tan^{2} (x) (1 + \frac{1}{sec ( x )}) - 1) = 0

각 부분을 개별적으로 해결

tan^{2} (x) (1 + \frac{1}{sec ( x )}) + 1 = 0 or tan^{2} (x) (1 + \frac{1}{sec ( x )}) - 1 = 0

tan^{2} (x) (1 + \frac{1}{sec ( x )}) + 1 = 0 :

해결책 없음

tan^{2} (x) (1 + \frac{1}{sec ( x )}) + 1 = 0

삼각성을 사용하여 다시 쓰기

1 + (1 + \frac{1}{sec ( x )}) tan^{2} (x)

피타고라스 정체성 사용:

tan^{2} (x) + 1 = sec^{2} (x)

tan^{2} (x) = sec^{2} (x) - 1

= 1 + (1 + \frac{1}{sec ( x )}) (sec^{2} (x) - 1)

1 + (1 + \frac{1}{sec ( x )}) (sec^{2} (x) - 1)

간소화하다 :

sec^{2} (x) + sec (x) - \frac{1}{sec ( x )}

1 + (1 + \frac{1}{sec ( x )}) (sec^{2} (x) - 1)

(1 + \frac{1}{sec ( x )}) (sec^{2} (x) - 1)

확대한다:

sec^{2} (x) - 1 + sec (x) - \frac{1}{sec ( x )}

(1 + \frac{1}{sec ( x )}) (sec^{2} (x) - 1)

호일 방법 적용:

(a + b) (c + d) = a c + a d + b c + b d

a = 1, b = \frac{1}{sec ( x )}, c = sec^{2} (x), d = - 1

= 1 \cdot sec^{2} (x) + 1 \cdot (- 1) + \frac{1}{sec ( x )} sec^{2} (x) + \frac{1}{sec ( x )} (- 1)

마이너스 플러스 규칙 적용

+ (- a) = - a

= 1 \cdot sec^{2} (x) - 1 \cdot 1 + \frac{1}{sec ( x )} sec^{2} (x) - 1 \cdot \frac{1}{sec ( x )}

1 \cdot sec^{2} (x) - 1 \cdot 1 + \frac{1}{sec ( x )} sec^{2} (x) - 1 \cdot \frac{1}{sec ( x )}

단순화하세요:

sec^{2} (x) - 1 + sec (x) - \frac{1}{sec ( x )}

1 \cdot sec^{2} (x) - 1 \cdot 1 + \frac{1}{sec ( x )} sec^{2} (x) - 1 \cdot \frac{1}{sec ( x )}

1 \cdot sec^{2} (x) = sec^{2} (x)

1 \cdot sec^{2} (x)

곱하다:

1 \cdot sec^{2} (x) = sec^{2} (x)

= sec^{2} (x)

1 \cdot 1 = 1

1 \cdot 1

숫자를 곱하시오:

1 \cdot 1 = 1

= 1

\frac{1}{sec ( x )} sec^{2} (x) = sec (x)

\frac{1}{sec ( x )} sec^{2} (x)

다중 분수:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{1 \cdot sec ^{2} ( x )}{sec ( x )}

곱하다:

1 \cdot sec^{2} (x) = sec^{2} (x)

= \frac{sec ^{2} ( x )}{sec ( x )}

공통 요인 취소:

sec (x)

= sec (x)

1 \cdot \frac{1}{sec ( x )} = \frac{1}{sec ( x )}

1 \cdot \frac{1}{sec ( x )}

곱하다:

1 \cdot \frac{1}{sec ( x )} = \frac{1}{sec ( x )}

= \frac{1}{sec ( x )}

= sec^{2} (x) - 1 + sec (x) - \frac{1}{sec ( x )}

= sec^{2} (x) - 1 + sec (x) - \frac{1}{sec ( x )}

= 1 + sec^{2} (x) - 1 + sec (x) - \frac{1}{sec ( x )}

1 + sec^{2} (x) - 1 + sec (x) - \frac{1}{sec ( x )}

단순화하세요:

sec^{2} (x) + sec (x) - \frac{1}{sec ( x )}

1 + sec^{2} (x) - 1 + sec (x) - \frac{1}{sec ( x )}

집단적 용어

= sec^{2} (x) + sec (x) - \frac{1}{sec ( x )} + 1 - 1

1 - 1 = 0

= sec^{2} (x) + sec (x) - \frac{1}{sec ( x )}

= sec^{2} (x) + sec (x) - \frac{1}{sec ( x )}

= sec^{2} (x) + sec (x) - \frac{1}{sec ( x )}

- \frac{1}{sec ( x )} + sec (x) + sec^{2} (x) = 0

대체로 해결

- \frac{1}{sec ( x )} + sec (x) + sec^{2} (x) = 0

하게:

sec (x) = u

- \frac{1}{u} + u + u^{2} = 0

- \frac{1}{u} + u + u^{2} = 0 : u \approx 0.75487 \dots

- \frac{1}{u} + u + u^{2} = 0

양쪽을 곱한 값

u

- \frac{1}{u} + u + u^{2} = 0

양쪽을 곱한 값

u

- \frac{1}{u} u + uu + u^{2} u = 0 \cdot u

단순화

- \frac{1}{u} u + uu + u^{2} u = 0 \cdot u

- \frac{1}{u} u

간소화하다 :

- 1

- \frac{1}{u} u

다중 분수:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= - \frac{1 \cdot u}{u}

공통 요인 취소:

u

= - 1

uu

간소화하다 :

u^{2}

uu

지수 규칙 적용:

a^{b} \cdot a^{c} = a^{b + c}

uu = u^{1 + 1}

= u^{1 + 1}

숫자 추가:

1 + 1 = 2

= u^{2}

u^{2} u

간소화하다 :

u^{3}

u^{2} u

지수 규칙 적용:

a^{b} \cdot a^{c} = a^{b + c}

u^{2} u = u^{2 + 1}

= u^{2 + 1}

숫자 추가:

2 + 1 = 3

= u^{3}

0 \cdot u

간소화하다 :

0

0 \cdot u

규칙 적용

0 \cdot a = 0

= 0

- 1 + u^{2} + u^{3} = 0

- 1 + u^{2} + u^{3} = 0

- 1 + u^{2} + u^{3} = 0

- 1 + u^{2} + u^{3} = 0

해결 :

u \approx 0.75487 \dots

- 1 + u^{2} + u^{3} = 0

표준 양식으로 작성

a_{n} x^{n} + \dots + a_{1} x + a_{0} = 0

u^{3} + u^{2} - 1 = 0

다음을 위한 하나의 솔루션 찾기

u^{3} + u^{2} - 1 = 0

뉴턴-랩슨을 이용하여:

u \approx 0.75487 \dots

u^{3} + u^{2} - 1 = 0

뉴턴-랩슨 근사 정의

f (u) = u^{3} + u^{2} - 1

f^{^{'}} (u)

찾다 :

3 u^{2} + 2 u

\frac{d}{d u} (u^{3} + u^{2} - 1)

합계/차이 규칙 적용:

(f \pm g)^{'} = f^{'} \pm g^{'}

= \frac{d}{d u} (u^{3}) + \frac{d}{d u} (u^{2}) - \frac{d}{d u} (1)

\frac{d}{d u} (u^{3}) = 3 u^{2}

\frac{d}{d u} (u^{3})

전원 규칙을 적용합니다:

\frac{d}{d x} (x^{a}) = a \cdot x^{a - 1}

= 3 u^{3 - 1}

단순화

= 3 u^{2}

\frac{d}{d u} (u^{2}) = 2 u

\frac{d}{d u} (u^{2})

전원 규칙을 적용합니다:

\frac{d}{d x} (x^{a}) = a \cdot x^{a - 1}

= 2 u^{2 - 1}

단순화

= 2 u

\frac{d}{d u} (1) = 0

\frac{d}{d u} (1)

상수의 도함수:

\frac{d}{d x} (a) = 0

= 0

= 3 u^{2} + 2 u - 0

단순화

= 3 u^{2} + 2 u

렛

u_{0} = 1

계산하다

u_{n + 1}

까지

Δ u_{n + 1} < 0.000001

u_{1} = 0.8 : Δ u_{1} = 0.2

f (u_{0}) = 1^{3} + 1^{2} - 1 = 1

f^{^{'}} (u_{0}) = 3 \cdot 1^{2} + 2 \cdot 1 = 5

u_{1} = 0.8

Δ u_{1} = ∣ 0.8 - 1 ∣ = 0.2

Δ u_{1} = 0.2

u_{2} = 0.75681 \dots : Δ u_{2} = 0.04318 \dots

f (u_{1}) = 0. 8^{3} + 0. 8^{2} - 1 = 0.152

f^{^{'}} (u_{1}) = 3 \cdot 0. 8^{2} + 2 \cdot 0.8 = 3.52

u_{2} = 0.75681 \dots

Δ u_{2} = ∣ 0.75681 \dots - 0.8 ∣ = 0.04318 \dots

Δ u_{2} = 0.04318 \dots

u_{3} = 0.75488 \dots : Δ u_{3} = 0.00193 \dots

f (u_{2}) = 0.75681 \dots^{3} + 0.75681 \dots^{2} - 1 = 0.00625 \dots

f^{^{'}} (u_{2}) = 3 \cdot 0.75681 \dots^{2} + 2 \cdot 0.75681 \dots = 3.23195 \dots

u_{3} = 0.75488 \dots

Δ u_{3} = ∣ 0.75488 \dots - 0.75681 \dots ∣ = 0.00193 \dots

Δ u_{3} = 0.00193 \dots

u_{4} = 0.75487 \dots : Δ u_{4} = 3.80818 E - 6

f (u_{3}) = 0.75488 \dots^{3} + 0.75488 \dots^{2} - 1 = 0.00001 \dots

f^{^{'}} (u_{3}) = 3 \cdot 0.75488 \dots^{2} + 2 \cdot 0.75488 \dots = 3.21930 \dots

u_{4} = 0.75487 \dots

Δ u_{4} = ∣ 0.75487 \dots - 0.75488 \dots ∣ = 3.80818 E - 6

Δ u_{4} = 3.80818 E - 6

u_{5} = 0.75487 \dots : Δ u_{5} = 1.47065 E - 11

f (u_{4}) = 0.75487 \dots^{3} + 0.75487 \dots^{2} - 1 = 4.73444 E - 11

f^{^{'}} (u_{4}) = 3 \cdot 0.75487 \dots^{2} + 2 \cdot 0.75487 \dots = 3.21927 \dots

u_{5} = 0.75487 \dots

Δ u_{5} = ∣ 0.75487 \dots - 0.75487 \dots ∣ = 1.47065 E - 11

Δ u_{5} = 1.47065 E - 11

u \approx 0.75487 \dots

긴 나눗셈 적용:

\frac{u ^{3} + u ^{2} - 1}{u - 0.75487 \dots} = u^{2} + 1.75487 \dots u + 1.32471 \dots

u^{2} + 1.75487 \dots u + 1.32471 \dots \approx 0

다음을 위한 하나의 솔루션 찾기

u^{2} + 1.75487 \dots u + 1.32471 \dots = 0

뉴턴-랩슨을 이용하여:

솔루션 없음

u \in R

u^{2} + 1.75487 \dots u + 1.32471 \dots = 0

뉴턴-랩슨 근사 정의

f (u) = u^{2} + 1.75487 \dots u + 1.32471 \dots

f^{^{'}} (u)

찾다 :

2 u + 1.75487 \dots

\frac{d}{d u} (u^{2} + 1.75487 \dots u + 1.32471 \dots)

합계/차이 규칙 적용:

(f \pm g)^{'} = f^{'} \pm g^{'}

= \frac{d}{d u} (u^{2}) + \frac{d}{d u} (1.75487 \dots u) + \frac{d}{d u} (1.32471 \dots)

\frac{d}{d u} (u^{2}) = 2 u

\frac{d}{d u} (u^{2})

전원 규칙을 적용합니다:

\frac{d}{d x} (x^{a}) = a \cdot x^{a - 1}

= 2 u^{2 - 1}

단순화

= 2 u

\frac{d}{d u} (1.75487 \dots u) = 1.75487 \dots

\frac{d}{d u} (1.75487 \dots u)

정수를 빼라:

(a \cdot f)^{'} = a \cdot f^{'}

= 1.75487 \dots \frac{d u}{d u}

공통 도함수 적용:

\frac{d u}{d u} = 1

= 1.75487 \dots \cdot 1

단순화

= 1.75487 \dots

\frac{d}{d u} (1.32471 \dots) = 0

\frac{d}{d u} (1.32471 \dots)

상수의 도함수:

\frac{d}{d x} (a) = 0

= 0

= 2 u + 1.75487 \dots + 0

단순화

= 2 u + 1.75487 \dots

렛

u_{0} = - 1

계산하다

u_{n + 1}

까지

Δ u_{n + 1} < 0.000001

u_{1} = 1.32471 \dots : Δ u_{1} = 2.32471 \dots

f (u_{0}) = (- 1)^{2} + 1.75487 \dots (- 1) + 1.32471 \dots = 0.56984 \dots

f^{^{'}} (u_{0}) = 2 (- 1) + 1.75487 \dots = - 0.24512 \dots

u_{1} = 1.32471 \dots

Δ u_{1} = ∣ 1.32471 \dots - (- 1) ∣ = 2.32471 \dots

Δ u_{1} = 2.32471 \dots

u_{2} = 0.09766 \dots : Δ u_{2} = 1.22705 \dots

f (u_{1}) = 1.32471 \dots^{2} + 1.75487 \dots \cdot 1.32471 \dots + 1.32471 \dots = 5.40431 \dots

f^{^{'}} (u_{1}) = 2 \cdot 1.32471 \dots + 1.75487 \dots = 4.40431 \dots

u_{2} = 0.09766 \dots

Δ u_{2} = ∣ 0.09766 \dots - 1.32471 \dots ∣ = 1.22705 \dots

Δ u_{2} = 1.22705 \dots

u_{3} = - 0.67437 \dots : Δ u_{3} = 0.77204 \dots

f (u_{2}) = 0.09766 \dots^{2} + 1.75487 \dots \cdot 0.09766 \dots + 1.32471 \dots = 1.50565 \dots

f^{^{'}} (u_{2}) = 2 \cdot 0.09766 \dots + 1.75487 \dots = 1.95021 \dots

u_{3} = - 0.67437 \dots

Δ u_{3} = ∣ - 0.67437 \dots - 0.09766 \dots ∣ = 0.77204 \dots

Δ u_{3} = 0.77204 \dots

u_{4} = - 2.14204 \dots : Δ u_{4} = 1.46766 \dots

f (u_{3}) = (- 0.67437 \dots)^{2} + 1.75487 \dots (- 0.67437 \dots) + 1.32471 \dots = 0.59605 \dots

f^{^{'}} (u_{3}) = 2 (- 0.67437 \dots) + 1.75487 \dots = 0.40612 \dots

u_{4} = - 2.14204 \dots

Δ u_{4} = ∣ - 2.14204 \dots - (- 0.67437 \dots) ∣ = 1.46766 \dots

Δ u_{4} = 1.46766 \dots

u_{5} = - 1.29037 \dots : Δ u_{5} = 0.85166 \dots

f (u_{4}) = (- 2.14204 \dots)^{2} + 1.75487 \dots (- 2.14204 \dots) + 1.32471 \dots = 2.15403 \dots

f^{^{'}} (u_{4}) = 2 (- 2.14204 \dots) + 1.75487 \dots = - 2.52920 \dots

u_{5} = - 1.29037 \dots

Δ u_{5} = ∣ - 1.29037 \dots - (- 2.14204 \dots) ∣ = 0.85166 \dots

Δ u_{5} = 0.85166 \dots

u_{6} = - 0.41210 \dots : Δ u_{6} = 0.87826 \dots

f (u_{5}) = (- 1.29037 \dots)^{2} + 1.75487 \dots (- 1.29037 \dots) + 1.32471 \dots = 0.72533 \dots

f^{^{'}} (u_{5}) = 2 (- 1.29037 \dots) + 1.75487 \dots = - 0.82587 \dots

u_{6} = - 0.41210 \dots

Δ u_{6} = ∣ - 0.41210 \dots - (- 1.29037 \dots) ∣ = 0.87826 \dots

Δ u_{6} = 0.87826 \dots

u_{7} = - 1.24093 \dots : Δ u_{7} = 0.82882 \dots

f (u_{6}) = (- 0.41210 \dots)^{2} + 1.75487 \dots (- 0.41210 \dots) + 1.32471 \dots = 0.77135 \dots

f^{^{'}} (u_{6}) = 2 (- 0.41210 \dots) + 1.75487 \dots = 0.93065 \dots

u_{7} = - 1.24093 \dots

Δ u_{7} = ∣ - 1.24093 \dots - (- 0.41210 \dots) ∣ = 0.82882 \dots

Δ u_{7} = 0.82882 \dots

u_{8} = - 0.29600 \dots : Δ u_{8} = 0.94492 \dots

f (u_{7}) = (- 1.24093 \dots)^{2} + 1.75487 \dots (- 1.24093 \dots) + 1.32471 \dots = 0.68694 \dots

f^{^{'}} (u_{7}) = 2 (- 1.24093 \dots) + 1.75487 \dots = - 0.72698 \dots

u_{8} = - 0.29600 \dots

Δ u_{8} = ∣ - 0.29600 \dots - (- 1.24093 \dots) ∣ = 0.94492 \dots

Δ u_{8} = 0.94492 \dots

u_{9} = - 1.06383 \dots : Δ u_{9} = 0.76782 \dots

f (u_{8}) = (- 0.29600 \dots)^{2} + 1.75487 \dots (- 0.29600 \dots) + 1.32471 \dots = 0.89288 \dots

f^{^{'}} (u_{8}) = 2 (- 0.29600 \dots) + 1.75487 \dots = 1.16286 \dots

u_{9} = - 1.06383 \dots

Δ u_{9} = ∣ - 1.06383 \dots - (- 0.29600 \dots) ∣ = 0.76782 \dots

Δ u_{9} = 0.76782 \dots

u_{10} = 0.51763 \dots : Δ u_{10} = 1.58147 \dots

f (u_{9}) = (- 1.06383 \dots)^{2} + 1.75487 \dots (- 1.06383 \dots) + 1.32471 \dots = 0.58956 \dots

f^{^{'}} (u_{9}) = 2 (- 1.06383 \dots) + 1.75487 \dots = - 0.37279 \dots

u_{10} = 0.51763 \dots

Δ u_{10} = ∣ 0.51763 \dots - (- 1.06383 \dots) ∣ = 1.58147 \dots

Δ u_{10} = 1.58147 \dots

해결 방법을 찾을 수 없습니다

해결책은

u \approx 0.75487 \dots

u \approx 0.75487 \dots

솔루션 확인

정의되지 않은 (특이점) 점 찾기:

u = 0

의 분모를 취하라

- \frac{1}{u} + u + u^{2}

그리고 0과 비교한다

u = 0

다음 지점은 정의되지 않았습니다

u = 0

정의되지 않은 점을 솔루션과 결합:

u \approx 0.75487 \dots

뒤로 대체

u = sec (x)

sec (x) \approx 0.75487 \dots

sec (x) \approx 0.75487 \dots

sec (x) = 0.75487 \dots :

해결책 없음

sec (x) = 0.75487 \dots

sec (x) \leq - 1 or sec (x) \geq 1

해결책 없음

모든 솔루션 결합

해결책 없음

tan^{2} (x) (1 + \frac{1}{sec ( x )}) - 1 = 0 : x = arcsec (1.24697 \dots) + 2 πn, x = 2 π - arcsec (1.24697 \dots) + 2 πn, x = arcsec (- 1.80193 \dots) + 2 πn, x = - arcsec (- 1.80193 \dots) + 2 πn

tan^{2} (x) (1 + \frac{1}{sec ( x )}) - 1 = 0

삼각성을 사용하여 다시 쓰기

- 1 + (1 + \frac{1}{sec ( x )}) tan^{2} (x)

피타고라스 정체성 사용:

tan^{2} (x) + 1 = sec^{2} (x)

tan^{2} (x) = sec^{2} (x) - 1

= - 1 + (1 + \frac{1}{sec ( x )}) (sec^{2} (x) - 1)

- 1 + (1 + \frac{1}{sec ( x )}) (sec^{2} (x) - 1)

간소화하다 :

sec^{2} (x) + sec (x) - \frac{1}{sec ( x )} - 2

- 1 + (1 + \frac{1}{sec ( x )}) (sec^{2} (x) - 1)

(1 + \frac{1}{sec ( x )}) (sec^{2} (x) - 1)

확대한다:

sec^{2} (x) - 1 + sec (x) - \frac{1}{sec ( x )}

(1 + \frac{1}{sec ( x )}) (sec^{2} (x) - 1)

호일 방법 적용:

(a + b) (c + d) = a c + a d + b c + b d

a = 1, b = \frac{1}{sec ( x )}, c = sec^{2} (x), d = - 1

= 1 \cdot sec^{2} (x) + 1 \cdot (- 1) + \frac{1}{sec ( x )} sec^{2} (x) + \frac{1}{sec ( x )} (- 1)

마이너스 플러스 규칙 적용

+ (- a) = - a

= 1 \cdot sec^{2} (x) - 1 \cdot 1 + \frac{1}{sec ( x )} sec^{2} (x) - 1 \cdot \frac{1}{sec ( x )}

1 \cdot sec^{2} (x) - 1 \cdot 1 + \frac{1}{sec ( x )} sec^{2} (x) - 1 \cdot \frac{1}{sec ( x )}

단순화하세요:

sec^{2} (x) - 1 + sec (x) - \frac{1}{sec ( x )}

1 \cdot sec^{2} (x) - 1 \cdot 1 + \frac{1}{sec ( x )} sec^{2} (x) - 1 \cdot \frac{1}{sec ( x )}

1 \cdot sec^{2} (x) = sec^{2} (x)

1 \cdot sec^{2} (x)

곱하다:

1 \cdot sec^{2} (x) = sec^{2} (x)

= sec^{2} (x)

1 \cdot 1 = 1

1 \cdot 1

숫자를 곱하시오:

1 \cdot 1 = 1

= 1

\frac{1}{sec ( x )} sec^{2} (x) = sec (x)

\frac{1}{sec ( x )} sec^{2} (x)

다중 분수:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{1 \cdot sec ^{2} ( x )}{sec ( x )}

곱하다:

1 \cdot sec^{2} (x) = sec^{2} (x)

= \frac{sec ^{2} ( x )}{sec ( x )}

공통 요인 취소:

sec (x)

= sec (x)

1 \cdot \frac{1}{sec ( x )} = \frac{1}{sec ( x )}

1 \cdot \frac{1}{sec ( x )}

곱하다:

1 \cdot \frac{1}{sec ( x )} = \frac{1}{sec ( x )}

= \frac{1}{sec ( x )}

= sec^{2} (x) - 1 + sec (x) - \frac{1}{sec ( x )}

= sec^{2} (x) - 1 + sec (x) - \frac{1}{sec ( x )}

= - 1 + sec^{2} (x) - 1 + sec (x) - \frac{1}{sec ( x )}

- 1 + sec^{2} (x) - 1 + sec (x) - \frac{1}{sec ( x )}

단순화하세요:

sec^{2} (x) + sec (x) - \frac{1}{sec ( x )} - 2

- 1 + sec^{2} (x) - 1 + sec (x) - \frac{1}{sec ( x )}

집단적 용어

= sec^{2} (x) + sec (x) - \frac{1}{sec ( x )} - 1 - 1

숫자를 빼세요:

- 1 - 1 = - 2

= sec^{2} (x) + sec (x) - \frac{1}{sec ( x )} - 2

= sec^{2} (x) + sec (x) - \frac{1}{sec ( x )} - 2

= sec^{2} (x) + sec (x) - \frac{1}{sec ( x )} - 2

- 2 - \frac{1}{sec ( x )} + sec (x) + sec^{2} (x) = 0

대체로 해결

- 2 - \frac{1}{sec ( x )} + sec (x) + sec^{2} (x) = 0

하게:

sec (x) = u

- 2 - \frac{1}{u} + u + u^{2} = 0

- 2 - \frac{1}{u} + u + u^{2} = 0 : u \approx - 0.44504 \dots, u \approx 1.24697 \dots, u \approx - 1.80193 \dots

- 2 - \frac{1}{u} + u + u^{2} = 0

양쪽을 곱한 값

u

- 2 - \frac{1}{u} + u + u^{2} = 0

양쪽을 곱한 값

u

- 2 u - \frac{1}{u} u + uu + u^{2} u = 0 \cdot u

단순화

- 2 u - \frac{1}{u} u + uu + u^{2} u = 0 \cdot u

- \frac{1}{u} u

간소화하다 :

- 1

- \frac{1}{u} u

다중 분수:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= - \frac{1 \cdot u}{u}

공통 요인 취소:

u

= - 1

uu

간소화하다 :

u^{2}

uu

지수 규칙 적용:

a^{b} \cdot a^{c} = a^{b + c}

uu = u^{1 + 1}

= u^{1 + 1}

숫자 추가:

1 + 1 = 2

= u^{2}

u^{2} u

간소화하다 :

u^{3}

u^{2} u

지수 규칙 적용:

a^{b} \cdot a^{c} = a^{b + c}

u^{2} u = u^{2 + 1}

= u^{2 + 1}

숫자 추가:

2 + 1 = 3

= u^{3}

0 \cdot u

간소화하다 :

0

0 \cdot u

규칙 적용

0 \cdot a = 0

= 0

- 2 u - 1 + u^{2} + u^{3} = 0

- 2 u - 1 + u^{2} + u^{3} = 0

- 2 u - 1 + u^{2} + u^{3} = 0

- 2 u - 1 + u^{2} + u^{3} = 0

해결 :

u \approx - 0.44504 \dots, u \approx 1.24697 \dots, u \approx - 1.80193 \dots

- 2 u - 1 + u^{2} + u^{3} = 0

표준 양식으로 작성

a_{n} x^{n} + \dots + a_{1} x + a_{0} = 0

u^{3} + u^{2} - 2 u - 1 = 0

다음을 위한 하나의 솔루션 찾기

u^{3} + u^{2} - 2 u - 1 = 0

뉴턴-랩슨을 이용하여:

u \approx - 0.44504 \dots

u^{3} + u^{2} - 2 u - 1 = 0

뉴턴-랩슨 근사 정의

f (u) = u^{3} + u^{2} - 2 u - 1

f^{^{'}} (u)

찾다 :

3 u^{2} + 2 u - 2

\frac{d}{d u} (u^{3} + u^{2} - 2 u - 1)

합계/차이 규칙 적용:

(f \pm g)^{'} = f^{'} \pm g^{'}

= \frac{d}{d u} (u^{3}) + \frac{d}{d u} (u^{2}) - \frac{d}{d u} (2 u) - \frac{d}{d u} (1)

\frac{d}{d u} (u^{3}) = 3 u^{2}

\frac{d}{d u} (u^{3})

전원 규칙을 적용합니다:

\frac{d}{d x} (x^{a}) = a \cdot x^{a - 1}

= 3 u^{3 - 1}

단순화

= 3 u^{2}

\frac{d}{d u} (u^{2}) = 2 u

\frac{d}{d u} (u^{2})

전원 규칙을 적용합니다:

\frac{d}{d x} (x^{a}) = a \cdot x^{a - 1}

= 2 u^{2 - 1}

단순화

= 2 u

\frac{d}{d u} (2 u) = 2

\frac{d}{d u} (2 u)

정수를 빼라:

(a \cdot f)^{'} = a \cdot f^{'}

= 2 \frac{d u}{d u}

공통 도함수 적용:

\frac{d u}{d u} = 1

= 2 \cdot 1

단순화

= 2

\frac{d}{d u} (1) = 0

\frac{d}{d u} (1)

상수의 도함수:

\frac{d}{d x} (a) = 0

= 0

= 3 u^{2} + 2 u - 2 - 0

단순화

= 3 u^{2} + 2 u - 2

렛

u_{0} = 0

계산하다

u_{n + 1}

까지

Δ u_{n + 1} < 0.000001

u_{1} = - 0.5 : Δ u_{1} = 0.5

f (u_{0}) = 0^{3} + 0^{2} - 2 \cdot 0 - 1 = - 1

f^{^{'}} (u_{0}) = 3 \cdot 0^{2} + 2 \cdot 0 - 2 = - 2

u_{1} = - 0.5

Δ u_{1} = ∣ - 0.5 - 0 ∣ = 0.5

Δ u_{1} = 0.5

u_{2} = - 0.44444 \dots : Δ u_{2} = 0.05555 \dots

f (u_{1}) = (- 0.5)^{3} + (- 0.5)^{2} - 2 (- 0.5) - 1 = 0.125

f^{^{'}} (u_{1}) = 3 (- 0.5)^{2} + 2 (- 0.5) - 2 = - 2.25

u_{2} = - 0.44444 \dots

Δ u_{2} = ∣ - 0.44444 \dots - (- 0.5) ∣ = 0.05555 \dots

Δ u_{2} = 0.05555 \dots

u_{3} = - 0.44504 \dots : Δ u_{3} = 0.00059 \dots

f (u_{2}) = (- 0.44444 \dots)^{3} + (- 0.44444 \dots)^{2} - 2 (- 0.44444 \dots) - 1 = - 0.00137 \dots

f^{^{'}} (u_{2}) = 3 (- 0.44444 \dots)^{2} + 2 (- 0.44444 \dots) - 2 = - 2.29629 \dots

u_{3} = - 0.44504 \dots

Δ u_{3} = ∣ - 0.44504 \dots - (- 0.44444 \dots) ∣ = 0.00059 \dots

Δ u_{3} = 0.00059 \dots

u_{4} = - 0.44504 \dots : Δ u_{4} = 5.19031 E - 8

f (u_{3}) = (- 0.44504 \dots)^{3} + (- 0.44504 \dots)^{2} - 2 (- 0.44504 \dots) - 1 = - 1.19164 E - 7

f^{^{'}} (u_{3}) = 3 (- 0.44504 \dots)^{2} + 2 (- 0.44504 \dots) - 2 = - 2.29589 \dots

u_{4} = - 0.44504 \dots

Δ u_{4} = ∣ - 0.44504 \dots - (- 0.44504 \dots) ∣ = 5.19031 E - 8

Δ u_{4} = 5.19031 E - 8

u \approx - 0.44504 \dots

긴 나눗셈 적용:

\frac{u ^{3} + u ^{2} - 2 u - 1}{u + 0.44504 \dots} = u^{2} + 0.55495 \dots u - 2.24697 \dots

u^{2} + 0.55495 \dots u - 2.24697 \dots \approx 0

다음을 위한 하나의 솔루션 찾기

u^{2} + 0.55495 \dots u - 2.24697 \dots = 0

뉴턴-랩슨을 이용하여:

u \approx 1.24697 \dots

u^{2} + 0.55495 \dots u - 2.24697 \dots = 0

뉴턴-랩슨 근사 정의

f (u) = u^{2} + 0.55495 \dots u - 2.24697 \dots

f^{^{'}} (u)

찾다 :

2 u + 0.55495 \dots

\frac{d}{d u} (u^{2} + 0.55495 \dots u - 2.24697 \dots)

합계/차이 규칙 적용:

(f \pm g)^{'} = f^{'} \pm g^{'}

= \frac{d}{d u} (u^{2}) + \frac{d}{d u} (0.55495 \dots u) - \frac{d}{d u} (2.24697 \dots)

\frac{d}{d u} (u^{2}) = 2 u

\frac{d}{d u} (u^{2})

전원 규칙을 적용합니다:

\frac{d}{d x} (x^{a}) = a \cdot x^{a - 1}

= 2 u^{2 - 1}

단순화

= 2 u

\frac{d}{d u} (0.55495 \dots u) = 0.55495 \dots

\frac{d}{d u} (0.55495 \dots u)

정수를 빼라:

(a \cdot f)^{'} = a \cdot f^{'}

= 0.55495 \dots \frac{d u}{d u}

공통 도함수 적용:

\frac{d u}{d u} = 1

= 0.55495 \dots \cdot 1

단순화

= 0.55495 \dots

\frac{d}{d u} (2.24697 \dots) = 0

\frac{d}{d u} (2.24697 \dots)

상수의 도함수:

\frac{d}{d x} (a) = 0

= 0

= 2 u + 0.55495 \dots - 0

단순화

= 2 u + 0.55495 \dots

렛

u_{0} = 4

계산하다

u_{n + 1}

까지

Δ u_{n + 1} < 0.000001

u_{1} = 2.13291 \dots : Δ u_{1} = 1.86708 \dots

f (u_{0}) = 4^{2} + 0.55495 \dots \cdot 4 - 2.24697 \dots = 15.97285 \dots

f^{^{'}} (u_{0}) = 2 \cdot 4 + 0.55495 \dots = 8.55495 \dots

u_{1} = 2.13291 \dots

Δ u_{1} = ∣ 2.13291 \dots - 4 ∣ = 1.86708 \dots

Δ u_{1} = 1.86708 \dots

u_{2} = 1.40979 \dots : Δ u_{2} = 0.72312 \dots

f (u_{1}) = 2.13291 \dots^{2} + 0.55495 \dots \cdot 2.13291 \dots - 2.24697 \dots = 3.48601 \dots

f^{^{'}} (u_{1}) = 2 \cdot 2.13291 \dots + 0.55495 \dots = 4.82078 \dots

u_{2} = 1.40979 \dots

Δ u_{2} = ∣ 1.40979 \dots - 2.13291 \dots ∣ = 0.72312 \dots

Δ u_{2} = 0.72312 \dots

u_{3} = 1.25483 \dots : Δ u_{3} = 0.15495 \dots

f (u_{2}) = 1.40979 \dots^{2} + 0.55495 \dots \cdot 1.40979 \dots - 2.24697 \dots = 0.52290 \dots

f^{^{'}} (u_{2}) = 2 \cdot 1.40979 \dots + 0.55495 \dots = 3.37453 \dots

u_{3} = 1.25483 \dots

Δ u_{3} = ∣ 1.25483 \dots - 1.40979 \dots ∣ = 0.15495 \dots

Δ u_{3} = 0.15495 \dots

u_{4} = 1.24699 \dots : Δ u_{4} = 0.00783 \dots

f (u_{3}) = 1.25483 \dots^{2} + 0.55495 \dots \cdot 1.25483 \dots - 2.24697 \dots = 0.02401 \dots

f^{^{'}} (u_{3}) = 2 \cdot 1.25483 \dots + 0.55495 \dots = 3.06462 \dots

u_{4} = 1.24699 \dots

Δ u_{4} = ∣ 1.24699 \dots - 1.25483 \dots ∣ = 0.00783 \dots

Δ u_{4} = 0.00783 \dots

u_{5} = 1.24697 \dots : Δ u_{5} = 0.00002 \dots

f (u_{4}) = 1.24699 \dots^{2} + 0.55495 \dots \cdot 1.24699 \dots - 2.24697 \dots = 0.00006 \dots

f^{^{'}} (u_{4}) = 2 \cdot 1.24699 \dots + 0.55495 \dots = 3.04895 \dots

u_{5} = 1.24697 \dots

Δ u_{5} = ∣ 1.24697 \dots - 1.24699 \dots ∣ = 0.00002 \dots

Δ u_{5} = 0.00002 \dots

u_{6} = 1.24697 \dots : Δ u_{6} = 1.32956 E - 10

f (u_{5}) = 1.24697 \dots^{2} + 0.55495 \dots \cdot 1.24697 \dots - 2.24697 \dots = 4.05373 E - 10

f^{^{'}} (u_{5}) = 2 \cdot 1.24697 \dots + 0.55495 \dots = 3.04891 \dots

u_{6} = 1.24697 \dots

Δ u_{6} = ∣ 1.24697 \dots - 1.24697 \dots ∣ = 1.32956 E - 10

Δ u_{6} = 1.32956 E - 10

u \approx 1.24697 \dots

긴 나눗셈 적용:

\frac{u ^{2} + 0.55495 \dots u - 2.24697 \dots}{u - 1.24697 \dots} = u + 1.80193 \dots

u + 1.80193 \dots \approx 0

u \approx - 1.80193 \dots

해결책은

u \approx - 0.44504 \dots, u \approx 1.24697 \dots, u \approx - 1.80193 \dots

u \approx - 0.44504 \dots, u \approx 1.24697 \dots, u \approx - 1.80193 \dots

솔루션 확인

정의되지 않은 (특이점) 점 찾기:

u = 0

의 분모를 취하라

- 2 - \frac{1}{u} + u + u^{2}

그리고 0과 비교한다

u = 0

다음 지점은 정의되지 않았습니다

u = 0

정의되지 않은 점을 솔루션과 결합:

u \approx - 0.44504 \dots, u \approx 1.24697 \dots, u \approx - 1.80193 \dots

뒤로 대체

u = sec (x)

sec (x) \approx - 0.44504 \dots, sec (x) \approx 1.24697 \dots, sec (x) \approx - 1.80193 \dots

sec (x) \approx - 0.44504 \dots, sec (x) \approx 1.24697 \dots, sec (x) \approx - 1.80193 \dots

sec (x) = - 0.44504 \dots :

해결책 없음

sec (x) = - 0.44504 \dots

sec (x) \leq - 1 or sec (x) \geq 1

해결책 없음

sec (x) = 1.24697 \dots : x = arcsec (1.24697 \dots) + 2 πn, x = 2 π - arcsec (1.24697 \dots) + 2 πn

sec (x) = 1.24697 \dots

트리거 역속성 적용

sec (x) = 1.24697 \dots

일반 솔루션

sec (x) = 1.24697 \dots

sec (x) = a \Rightarrow x = arcsec (a) + 2 πn, x = 2 π - arcsec (a) + 2 πn

x = arcsec (1.24697 \dots) + 2 πn, x = 2 π - arcsec (1.24697 \dots) + 2 πn

x = arcsec (1.24697 \dots) + 2 πn, x = 2 π - arcsec (1.24697 \dots) + 2 πn

sec (x) = - 1.80193 \dots : x = arcsec (- 1.80193 \dots) + 2 πn, x = - arcsec (- 1.80193 \dots) + 2 πn

sec (x) = - 1.80193 \dots

트리거 역속성 적용

sec (x) = - 1.80193 \dots

일반 솔루션

sec (x) = - 1.80193 \dots

sec (x) = - a \Rightarrow x = arcsec (- a) + 2 πn, x = - arcsec (- a) + 2 πn

x = arcsec (- 1.80193 \dots) + 2 πn, x = - arcsec (- 1.80193 \dots) + 2 πn

x = arcsec (- 1.80193 \dots) + 2 πn, x = - arcsec (- 1.80193 \dots) + 2 πn

모든 솔루션 결합

x = arcsec (1.24697 \dots) + 2 πn, x = 2 π - arcsec (1.24697 \dots) + 2 πn, x = arcsec (- 1.80193 \dots) + 2 πn, x = - arcsec (- 1.80193 \dots) + 2 πn

모든 솔루션 결합

x = arcsec (1.24697 \dots) + 2 πn, x = 2 π - arcsec (1.24697 \dots) + 2 πn, x = arcsec (- 1.80193 \dots) + 2 πn, x = - arcsec (- 1.80193 \dots) + 2 πn

해법을 원래 방정식에 연결하여 검증

솔루션을 에 연결하여 확인합니다

tan^{2} (x) = \frac{1}{cos ( x ) + 1}

방정식에 맞지 않는 것은 제거하십시오.

솔루션 확인

arcsec (1.24697 \dots) + 2 πn :

참

arcsec (1.24697 \dots) + 2 πn

n = 1

끼우다

arcsec (1.24697 \dots) + 2 π 1

tan^{2} (x) = \frac{1}{cos ( x ) + 1}

위한 {\ quad}끼우다{\ quad}

x = arcsec (1.24697 \dots) + 2 π 1

tan^{2} (arcsec (1.24697 \dots) + 2 π 1) = \frac{1}{cos ( arcsec ( 1.24697 \dots ) + 2 π 1 ) + 1}

다듬다

0.55495 \dots = 0.55495 \dots

\Rightarrow 참

솔루션 확인

2 π - arcsec (1.24697 \dots) + 2 πn :

참

2 π - arcsec (1.24697 \dots) + 2 πn

n = 1

끼우다

2 π - arcsec (1.24697 \dots) + 2 π 1

tan^{2} (x) = \frac{1}{cos ( x ) + 1}

위한 {\ quad}끼우다{\ quad}

x = 2 π - arcsec (1.24697 \dots) + 2 π 1

tan^{2} (2 π - arcsec (1.24697 \dots) + 2 π 1) = \frac{1}{cos ( 2 π - arcsec ( 1.24697 \dots ) + 2 π 1 ) + 1}

다듬다

0.55495 \dots = 0.55495 \dots

\Rightarrow 참

솔루션 확인

arcsec (- 1.80193 \dots) + 2 πn :

참

arcsec (- 1.80193 \dots) + 2 πn

n = 1

끼우다

arcsec (- 1.80193 \dots) + 2 π 1

tan^{2} (x) = \frac{1}{cos ( x ) + 1}

위한 {\ quad}끼우다{\ quad}

x = arcsec (- 1.80193 \dots) + 2 π 1

tan^{2} (arcsec (- 1.80193 \dots) + 2 π 1) = \frac{1}{cos ( arcsec ( - 1.80193 \dots ) + 2 π 1 ) + 1}

다듬다

2.24697 \dots = 2.24697 \dots

\Rightarrow 참

솔루션 확인

- arcsec (- 1.80193 \dots) + 2 πn :

참

- arcsec (- 1.80193 \dots) + 2 πn

n = 1

끼우다

- arcsec (- 1.80193 \dots) + 2 π 1

tan^{2} (x) = \frac{1}{cos ( x ) + 1}

위한 {\ quad}끼우다{\ quad}

x = - arcsec (- 1.80193 \dots) + 2 π 1

tan^{2} (- arcsec (- 1.80193 \dots) + 2 π 1) = \frac{1}{cos ( - arcsec ( - 1.80193 \dots ) + 2 π 1 ) + 1}

다듬다

2.24697 \dots = 2.24697 \dots

\Rightarrow 참

x = arcsec (1.24697 \dots) + 2 πn, x = 2 π - arcsec (1.24697 \dots) + 2 πn, x = arcsec (- 1.80193 \dots) + 2 πn, x = - arcsec (- 1.80193 \dots) + 2 πn

해를 10진수 형식으로 표시

x = 0.64026 \dots + 2 πn, x = 2 π - 0.64026 \dots + 2 πn, x = 2.15910 \dots + 2 πn, x = - 2.15910 \dots + 2 πn

그래프

인기 있는 예

(sin^2(x)-2cos(x)+1)/4 =0

\frac{sin ^{2} ( x ) - 2 cos ( x ) + 1}{4} = 0

cos^2(x)+cos^4(x)+cos^6(x)=0

cos^{2} (x) + cos^{4} (x) + cos^{6} (x) = 0

sin (x) = \frac{- 1}{4}

sin^4(x)-sin^2(x)=0

sin^{4} (x) - sin^{2} (x) = 0

(cos(t)-4)(2sin^2(t)-1)=0

(cos (t) - 4) (2 sin^{2} (t) - 1) = 0