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受欢迎的三角函数 >

cos(θ)(cos(θ)-2)=1,-180<= θ<= 180

解答

cos (θ) (cos (θ) - 2) = 1, - 18 0^{\circ} \leq θ \leq 18 0^{\circ}

解答

θ = 1.99787 \dots, θ = - 1.99787 \dots

求解步骤

cos (θ) (cos (θ) - 2) = 1, - 18 0^{\circ} \leq θ \leq 18 0^{\circ}

用替代法求解

cos (θ) (cos (θ) - 2) = 1

令

: cos (θ) = u

u (u - 2) = 1

u (u - 2) = 1 : u = 1 + 2, u = 1 - 2

u (u - 2) = 1

展开

u (u - 2) : u^{2} - 2 u

u (u - 2)

使用分配律:

a (b - c) = ab - a c

a = u, b = u, c = 2

= uu - u \cdot 2

= uu - 2 u

uu = u^{2}

uu

使用指数法则:

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

uu = u^{1 + 1}

= u^{1 + 1}

数字相加:

1 + 1 = 2

= u^{2}

= u^{2} - 2 u

u^{2} - 2 u = 1

将

1

para o lado esquerdo

u^{2} - 2 u = 1

两边减去

1

u^{2} - 2 u - 1 = 1 - 1

化简

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

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

使用求根公式求解

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

二次方程求根公式:

若

a = 1, b = - 2, c = - 1

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

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

(- 2)^{2} - 4 \cdot 1 \cdot (- 1) = 22

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

使用法则

- (- a) = a

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

使用指数法则:

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

若

n

是偶数

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

= 2^{2} + 4 \cdot 1 \cdot 1

数字相乘:

4 \cdot 1 \cdot 1 = 4

= 2^{2} + 4

2^{2} = 4

= 4 + 4

数字相加:

4 + 4 = 8

= 8

8

质因数分解:

2^{3}

8

8

除以

28 = 4 \cdot 2

= 2 \cdot 4

4

除以

24 = 2 \cdot 2

= 2 \cdot 2 \cdot 2

2

是质数，因此无法进一步因数分解

= 2 \cdot 2 \cdot 2

= 2^{3}

= 2^{3}

使用指数法则:

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

= 2^{2} \cdot 2

使用根式运算法则:

n ab = n a n b

= 2 2^{2}

使用根式运算法则:

n a^{n} = a

2^{2} = 2

= 22

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

将解分隔开

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

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

\frac{- ( - 2 ) + 2 2}{2 \cdot 1}

使用法则

- (- a) = a

= \frac{2 + 2 2}{2 \cdot 1}

数字相乘:

2 \cdot 1 = 2

= \frac{2 + 2 2}{2}

分解

2 + 22 : 2 (1 + 2)

2 + 22

改写为

= 2 \cdot 1 + 22

因式分解出通项

2

= 2 (1 + 2)

= \frac{2 ( 1 + 2 )}{2}

数字相除:

\frac{2}{2} = 1

= 1 + 2

u = \frac{- ( - 2 ) - 2 2}{2 \cdot 1} : 1 - 2

\frac{- ( - 2 ) - 2 2}{2 \cdot 1}

使用法则

- (- a) = a

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

数字相乘:

2 \cdot 1 = 2

= \frac{2 - 2 2}{2}

分解

2 - 22 : 2 (1 - 2)

2 - 22

改写为

= 2 \cdot 1 - 22

因式分解出通项

2

= 2 (1 - 2)

= \frac{2 ( 1 - 2 )}{2}

数字相除:

\frac{2}{2} = 1

= 1 - 2

二次方程组的解是:

u = 1 + 2, u = 1 - 2

u = cos (θ)

代回

cos (θ) = 1 + 2, cos (θ) = 1 - 2

cos (θ) = 1 + 2, cos (θ) = 1 - 2

cos (θ) = 1 + 2, - 18 0^{\circ} \leq θ \leq 18 0^{\circ} :

无解

cos (θ) = 1 + 2, - 18 0^{\circ} \leq θ \leq 18 0^{\circ}

- 1 \leq cos (x) \leq 1

无解

cos (θ) = 1 - 2, - 18 0^{\circ} \leq θ \leq 18 0^{\circ} : θ = arccos (1 - 2), θ = - arccos (1 - 2)

cos (θ) = 1 - 2, - 18 0^{\circ} \leq θ \leq 18 0^{\circ}

使用反三角函数性质

cos (θ) = 1 - 2

cos (θ) = 1 - 2

的通解

cos (x) = - a \Rightarrow x = arccos (- a) + 36 0^{\circ} n, x = - arccos (- a) + 36 0^{\circ} n

θ = arccos (1 - 2) + 36 0^{\circ} n, θ = - arccos (1 - 2) + 36 0^{\circ} n

θ = arccos (1 - 2) + 36 0^{\circ} n, θ = - arccos (1 - 2) + 36 0^{\circ} n

在

- 18 0^{\circ} \leq θ \leq 18 0^{\circ}

范围内的解

θ = arccos (1 - 2), θ = - arccos (1 - 2)

合并所有解

θ = arccos (1 - 2), θ = - arccos (1 - 2)

以小数形式表示解

θ = 1.99787 \dots, θ = - 1.99787 \dots

作图

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