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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 = 2.23703 \dots + 2 πn, θ = - 2.23703 \dots + 2 πn

+1

度数

θ = 128.17270 \dots^{\circ} + 36 0^{\circ} n, θ = - 128.17270 \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

- (- a) = a

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

(- 1)^{2} = 1

(- 1)^{2}

使用指数法则:

(- a)^{n} = a^{n},

若

n

是偶数

(- 1)^{2} = 1^{2}

= 1^{2}

使用法则

1^{a} = 1

= 1

4 \cdot 1 \cdot 1 = 4

4 \cdot 1 \cdot 1

数字相乘:

4 \cdot 1 \cdot 1 = 4

= 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}

使用法则

- (- a) = a

= \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}

使用法则

- (- a) = a

= \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} :

无解

cos (θ) = \frac{1 + 5}{2}

- 1 \leq cos (x) \leq 1

无解

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

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

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

合并所有解

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

以小数形式表示解

θ = 2.23703 \dots + 2 πn, θ = - 2.23703 \dots + 2 πn

作图

流行的例子

3cot^2(x)-4csc(x)=1

3 cot^{2} (x) - 4 csc (x) = 1

cos(x)=sin(2x)+cos(3x)

cos (x) = sin (2 x) + cos (3 x)

tan(x)sec(x)-2tan(x)=0

tan (x) sec (x) - 2 tan (x) = 0

-9cos^2(θ)+9=17sin(θ)-8

- 9 cos^{2} (θ) + 9 = 17 sin (θ) - 8

4 tan^{2} (θ) - 9 = 0