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受欢迎的三角函数 >

5cos^2(x)+cos(x)=0

解答

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

解答

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn, x = 1.77215 \dots + 2 πn, x = - 1.77215 \dots + 2 πn

+1

度数

x = 9 0^{\circ} + 36 0^{\circ} n, x = 27 0^{\circ} + 36 0^{\circ} n, x = 101.53695 \dots^{\circ} + 36 0^{\circ} n, x = - 101.53695 \dots^{\circ} + 36 0^{\circ} n

求解步骤

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

用替代法求解

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

令

: cos (x) = u

5 u^{2} + u = 0

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

5 u^{2} + u = 0

使用求根公式求解

5 u^{2} + u = 0

二次方程求根公式:

若

a = 5, b = 1, c = 0

u_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 5 \cdot 0}{2 \cdot 5}

u_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 5 \cdot 0}{2 \cdot 5}

1^{2} - 4 \cdot 5 \cdot 0 = 1

1^{2} - 4 \cdot 5 \cdot 0

使用法则

1^{a} = 1

1^{2} = 1

= 1 - 4 \cdot 5 \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 5}

将解分隔开

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

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

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

数字相加/相减:

- 1 + 1 = 0

= \frac{0}{2 \cdot 5}

数字相乘:

2 \cdot 5 = 10

= \frac{0}{10}

使用法则

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

= 0

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

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

数字相减:

- 1 - 1 = - 2

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

数字相乘:

2 \cdot 5 = 10

= \frac{- 2}{10}

使用分式法则:

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

= - \frac{2}{10}

约分:

2

= - \frac{1}{5}

二次方程组的解是:

u = 0, u = - \frac{1}{5}

u = cos (x)

代回

cos (x) = 0, cos (x) = - \frac{1}{5}

cos (x) = 0, cos (x) = - \frac{1}{5}

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

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

使用反三角函数性质

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

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

的通解

cos (x) = - a \Rightarrow x = arccos (- a) + 2 πn, x = - arccos (- a) + 2 πn

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

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

合并所有解

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn, x = arccos (- \frac{1}{5}) + 2 πn, x = - arccos (- \frac{1}{5}) + 2 πn

以小数形式表示解

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn, x = 1.77215 \dots + 2 πn, x = - 1.77215 \dots + 2 πn

作图

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