zs

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

受欢迎的三角函数 >

cos^2(θ)+cos(θ)-1=0

解答

cos^{2} (θ) + cos (θ) - 1 = 0

解答

θ = 0.90455 \dots + 2 πn, θ = 2 π - 0.90455 \dots + 2 πn

+1

度数

θ = 51.82729 \dots^{\circ} + 36 0^{\circ} n, θ = 308.17270 \dots^{\circ} + 36 0^{\circ} n

求解步骤

cos^{2} (θ) + cos (θ) - 1 = 0

用替代法求解

cos^{2} (θ) + cos (θ) - 1 = 0

令

: cos (θ) = u

u^{2} + u - 1 = 0

u^{2} + u - 1 = 0 : u = \frac{- 1 + 5}{2}, u = \frac{- 1 - 5}{2}

u^{2} + u - 1 = 0

使用求根公式求解

u^{2} + u - 1 = 0

二次方程求根公式:

若

a = 1, b = 1, c = - 1

u_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 1 \cdot ( - 1 )}{2 \cdot 1}

u_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 1 \cdot ( - 1 )}{2 \cdot 1}

1^{2} - 4 \cdot 1 \cdot (- 1) = 5

1^{2} - 4 \cdot 1 \cdot (- 1)

使用法则

1^{a} = 1

1^{2} = 1

= 1 - 4 \cdot 1 \cdot (- 1)

使用法则

- (- a) = a

= 1 + 4 \cdot 1 \cdot 1

数字相乘:

4 \cdot 1 \cdot 1 = 4

= 1 + 4

数字相加:

1 + 4 = 5

= 5

u_{1, 2} = \frac{- 1 \pm 5}{2 \cdot 1}

将解分隔开

u_{1} = \frac{- 1 + 5}{2 \cdot 1}, u_{2} = \frac{- 1 - 5}{2 \cdot 1}

u = \frac{- 1 + 5}{2 \cdot 1} : \frac{- 1 + 5}{2}

\frac{- 1 + 5}{2 \cdot 1}

数字相乘:

2 \cdot 1 = 2

= \frac{- 1 + 5}{2}

u = \frac{- 1 - 5}{2 \cdot 1} : \frac{- 1 - 5}{2}

\frac{- 1 - 5}{2 \cdot 1}

数字相乘:

2 \cdot 1 = 2

= \frac{- 1 - 5}{2}

二次方程组的解是:

u = \frac{- 1 + 5}{2}, u = \frac{- 1 - 5}{2}

u = cos (θ)

代回

cos (θ) = \frac{- 1 + 5}{2}, cos (θ) = \frac{- 1 - 5}{2}

cos (θ) = \frac{- 1 + 5}{2}, cos (θ) = \frac{- 1 - 5}{2}

cos (θ) = \frac{- 1 + 5}{2} : θ = arccos (\frac{- 1 + 5}{2}) + 2 πn, θ = 2 π - arccos (\frac{- 1 + 5}{2}) + 2 πn

cos (θ) = \frac{- 1 + 5}{2}

使用反三角函数性质

cos (θ) = \frac{- 1 + 5}{2}

cos (θ) = \frac{- 1 + 5}{2}

的通解

cos (x) = a \Rightarrow x = arccos (a) + 2 πn, x = 2 π - arccos (a) + 2 πn

θ = arccos (\frac{- 1 + 5}{2}) + 2 πn, θ = 2 π - arccos (\frac{- 1 + 5}{2}) + 2 πn

θ = arccos (\frac{- 1 + 5}{2}) + 2 πn, θ = 2 π - arccos (\frac{- 1 + 5}{2}) + 2 πn

cos (θ) = \frac{- 1 - 5}{2} :

无解

cos (θ) = \frac{- 1 - 5}{2}

- 1 \leq cos (x) \leq 1

无解

合并所有解

θ = arccos (\frac{- 1 + 5}{2}) + 2 πn, θ = 2 π - arccos (\frac{- 1 + 5}{2}) + 2 πn

以小数形式表示解

θ = 0.90455 \dots + 2 πn, θ = 2 π - 0.90455 \dots + 2 πn

作图

流行的例子

sin^{2} (x) - 3 = 0

cos(θ)csc(θ)=-sqrt(2)cos(θ)

cos (θ) csc (θ) = - 2 cos (θ)

csc^2(x)-4cot(x)=-2

csc^{2} (x) - 4 cot (x) = - 2

sin(2x-pi/3)=(sqrt(3))/2

sin (2 x - \frac{π}{3}) = \frac{3}{2}

-sin^3(t)+sin(2t)cos(t)=0

- sin^{3} (t) + sin (2 t) cos (t) = 0