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受欢迎的三角函数 >

cos^2(x)+2cos(x)=sin^2(x/2)-2

解答

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

解答

x = π + 2 πn

+1

度数

x = 18 0^{\circ} + 36 0^{\circ} n

求解步骤

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

两边减去

sin^{2} (\frac{x}{2}) - 2

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

使用三角恒等式改写

2 + cos^{2} (x) - sin^{2} (\frac{x}{2}) + 2 cos (x)

利用以下特性:

sin^{2} (θ) = \frac{1 - cos ( 2 θ )}{2}

使用倍角公式

cos (2 θ) = 1 - 2 sin^{2} (θ)

交换两边

2 sin^{2} (θ) - 1 = - cos (2 θ)

两边加上

1

2 sin^{2} (θ) = 1 - cos (2 θ)

两边除以

2

sin^{2} (θ) = \frac{1 - cos ( 2 θ )}{2}

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

化简

2 + cos^{2} (x) - \frac{1 - cos ( 2 \cdot \frac{x}{2} )}{2} + 2 cos (x) : \frac{5 cos ( x ) + 2 cos ^{2} ( x ) - 1}{2} + 2

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

乘

2 \cdot \frac{x}{2} : x

2 \cdot \frac{x}{2}

分式相乘:

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

= \frac{x \cdot 2}{2}

约分:

2

= x

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

合并分式

- \frac{- cos ( x ) + 1}{2} + cos^{2} (x) + 2 cos (x) : \frac{5 cos ( x ) + 2 cos ^{2} ( x ) - 1}{2}

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

将项转换为分式:

cos^{2} (x) = \frac{cos ^{2} ( x ) 2}{2}, 2 cos (x) = \frac{2 cos ( x ) 2}{2}

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

因为分母相等，所以合并分式:

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

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

数字相乘:

2 \cdot 2 = 4

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

乘开

- (1 - cos (x)) + cos^{2} (x) \cdot 2 + 4 cos (x) : 5 cos (x) + 2 cos^{2} (x) - 1

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

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

- (1 - cos (x)) : - 1 + cos (x)

- (1 - cos (x))

打开括号

= - (1) - (- cos (x))

使用加减运算法则

- (- a) = a, - (a) = - a

= - 1 + cos (x)

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

化简

- 1 + cos (x) + cos^{2} (x) \cdot 2 + 4 cos (x) : 5 cos (x) + 2 cos^{2} (x) - 1

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

对同类项分组

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

同类项相加:

cos (x) + 4 cos (x) = 5 cos (x)

= 5 cos (x) + 2 cos^{2} (x) - 1

= 5 cos (x) + 2 cos^{2} (x) - 1

= \frac{5 cos ( x ) + 2 cos ^{2} ( x ) - 1}{2}

= \frac{2 cos ^{2} ( x ) + 5 cos ( x ) - 1}{2} + 2

= \frac{5 cos ( x ) + 2 cos ^{2} ( x ) - 1}{2} + 2

2 + \frac{- 1 + 2 cos ^{2} ( x ) + 5 cos ( x )}{2} = 0

用替代法求解

2 + \frac{- 1 + 2 cos ^{2} ( x ) + 5 cos ( x )}{2} = 0

令

: cos (x) = u

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

2 + \frac{- 1 + 2 u ^{2} + 5 u}{2} = 0 : u = - 1, u = - \frac{3}{2}

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

在两边乘以

2

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

在两边乘以

2

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

化简

4 - 1 + 2 u^{2} + 5 u = 0

4 - 1 + 2 u^{2} + 5 u = 0

2 u^{2} + 5 u + 3 = 0

使用求根公式求解

2 u^{2} + 5 u + 3 = 0

二次方程求根公式:

若

a = 2, b = 5, c = 3

u_{1, 2} = \frac{- 5 \pm 5 ^{2} - 4 \cdot 2 \cdot 3}{2 \cdot 2}

u_{1, 2} = \frac{- 5 \pm 5 ^{2} - 4 \cdot 2 \cdot 3}{2 \cdot 2}

5^{2} - 4 \cdot 2 \cdot 3 = 1

5^{2} - 4 \cdot 2 \cdot 3

数字相乘:

4 \cdot 2 \cdot 3 = 24

= 5^{2} - 24

5^{2} = 25

= 25 - 24

数字相减:

25 - 24 = 1

= 1

使用法则

1 = 1

= 1

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

将解分隔开

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

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

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

数字相加/相减:

- 5 + 1 = - 4

= \frac{- 4}{2 \cdot 2}

数字相乘:

2 \cdot 2 = 4

= \frac{- 4}{4}

使用分式法则:

\frac{- a}{b} = - \frac{a}{b}

= - \frac{4}{4}

使用法则

\frac{a}{a} = 1

= - 1

u = \frac{- 5 - 1}{2 \cdot 2} : - \frac{3}{2}

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

数字相减:

- 5 - 1 = - 6

= \frac{- 6}{2 \cdot 2}

数字相乘:

2 \cdot 2 = 4

= \frac{- 6}{4}

使用分式法则:

\frac{- a}{b} = - \frac{a}{b}

= - \frac{6}{4}

约分:

2

= - \frac{3}{2}

二次方程组的解是:

u = - 1, u = - \frac{3}{2}

u = cos (x)

代回

cos (x) = - 1, cos (x) = - \frac{3}{2}

cos (x) = - 1, cos (x) = - \frac{3}{2}

cos (x) = - 1 : x = π + 2 πn

cos (x) = - 1

cos (x) = - 1

的通解

cos (x)

周期表（周期为

2 πn

）：

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

x = π + 2 πn

x = π + 2 πn

cos (x) = - \frac{3}{2} :

无解

cos (x) = - \frac{3}{2}

- 1 \leq cos (x) \leq 1

无解

合并所有解

x = π + 2 πn

作图

流行的例子

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

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

cos^2(x)-sin^2(x)=1-2sin(x)

cos^{2} (x) - sin^{2} (x) = 1 - 2 sin (x)

cos (x) = - 0.34

2sin^2(x)=3sin(x)-1

2 sin^{2} (x) = 3 sin (x) - 1

2cos(x)sin(x)-sin(x)=0

2 cos (x) sin (x) - sin (x) = 0