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受欢迎的三角函数 >

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

解答

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

解答

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn, x = \frac{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn

+1

度数

x = 9 0^{\circ} + 36 0^{\circ} n, x = 27 0^{\circ} + 36 0^{\circ} n, x = 12 0^{\circ} + 36 0^{\circ} n, x = 24 0^{\circ} + 36 0^{\circ} n

求解步骤

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

化简

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

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

将项转换为分式:

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

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

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

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

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

使用三角恒等式改写

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

使用倍角公式:

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

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

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) + cos (x)

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

用替代法求解

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

令

: cos (x) = u

u + 2 u^{2} = 0

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

u + 2 u^{2} = 0

改写成标准形式

a x^{2} + b x + c = 0

2 u^{2} + u = 0

使用求根公式求解

2 u^{2} + u = 0

二次方程求根公式:

若

a = 2, b = 1, c = 0

u_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 2 \cdot 0}{2 \cdot 2}

u_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 2 \cdot 0}{2 \cdot 2}

1^{2} - 4 \cdot 2 \cdot 0 = 1

1^{2} - 4 \cdot 2 \cdot 0

使用法则

1^{a} = 1

1^{2} = 1

= 1 - 4 \cdot 2 \cdot 0

使用法则

0 \cdot a = 0

= 1 - 0

数字相减:

1 - 0 = 1

= 1

使用法则

1 = 1

= 1

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

将解分隔开

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

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

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

数字相加/相减:

- 1 + 1 = 0

= \frac{0}{2 \cdot 2}

数字相乘:

2 \cdot 2 = 4

= \frac{0}{4}

使用法则

\frac{0}{a} = 0, a \neq = 0

= 0

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

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

数字相减:

- 1 - 1 = - 2

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

数字相乘:

2 \cdot 2 = 4

= \frac{- 2}{4}

使用分式法则:

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

= - \frac{2}{4}

约分:

2

= - \frac{1}{2}

二次方程组的解是:

u = 0, u = - \frac{1}{2}

u = cos (x)

代回

cos (x) = 0, cos (x) = - \frac{1}{2}

cos (x) = 0, cos (x) = - \frac{1}{2}

cos (x) = 0 : x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

cos (x) = 0

cos (x) = 0

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 = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

cos (x) = - \frac{1}{2} : x = \frac{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn

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

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

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 = \frac{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn

x = \frac{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn

合并所有解

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn, x = \frac{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn

作图

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