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受欢迎的三角函数 >

7cos^2(x)+14cos(x)+7=0

解答

7 cos^{2} (x) + 14 cos (x) + 7 = 0

解答

x = π + 2 πn

+1

度数

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

求解步骤

7 cos^{2} (x) + 14 cos (x) + 7 = 0

用替代法求解

7 cos^{2} (x) + 14 cos (x) + 7 = 0

令

: cos (x) = u

7 u^{2} + 14 u + 7 = 0

7 u^{2} + 14 u + 7 = 0 : u = - 1

7 u^{2} + 14 u + 7 = 0

使用求根公式求解

7 u^{2} + 14 u + 7 = 0

二次方程求根公式:

若

a = 7, b = 14, c = 7

u_{1, 2} = \frac{- 14 \pm 1 4 ^{2} - 4 \cdot 7 \cdot 7}{2 \cdot 7}

u_{1, 2} = \frac{- 14 \pm 1 4 ^{2} - 4 \cdot 7 \cdot 7}{2 \cdot 7}

1 4^{2} - 4 \cdot 7 \cdot 7 = 0

1 4^{2} - 4 \cdot 7 \cdot 7

数字相乘:

4 \cdot 7 \cdot 7 = 196

= 1 4^{2} - 196

1 4^{2} = 196

= 196 - 196

数字相减:

196 - 196 = 0

= 0

u_{1, 2} = \frac{- 14 \pm 0}{2 \cdot 7}

u = \frac{- 14}{2 \cdot 7}

\frac{- 14}{2 \cdot 7} = - 1

\frac{- 14}{2 \cdot 7}

数字相乘:

2 \cdot 7 = 14

= \frac{- 14}{14}

使用分式法则:

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

= - \frac{14}{14}

使用法则

\frac{a}{a} = 1

= - 1

u = - 1

二次方程组的解是:

u = - 1

u = cos (x)

代回

cos (x) = - 1

cos (x) = - 1

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

合并所有解

x = π + 2 πn

作图

流行的例子

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