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受欢迎的三角函数 >

-20cos^2(x)-36cos(x)=16

解答

- 20 cos^{2} (x) - 36 cos (x) = 16

解答

x = π + 2 πn, x = 2.49809 \dots + 2 πn, x = - 2.49809 \dots + 2 πn

+1

度数

x = 18 0^{\circ} + 36 0^{\circ} n, x = 143.13010 \dots^{\circ} + 36 0^{\circ} n, x = - 143.13010 \dots^{\circ} + 36 0^{\circ} n

求解步骤

- 20 cos^{2} (x) - 36 cos (x) = 16

用替代法求解

- 20 cos^{2} (x) - 36 cos (x) = 16

令

: cos (x) = u

- 20 u^{2} - 36 u = 16

- 20 u^{2} - 36 u = 16 : u = - 1, u = - \frac{4}{5}

- 20 u^{2} - 36 u = 16

将

16

para o lado esquerdo

- 20 u^{2} - 36 u = 16

两边减去

16

- 20 u^{2} - 36 u - 16 = 16 - 16

化简

- 20 u^{2} - 36 u - 16 = 0

- 20 u^{2} - 36 u - 16 = 0

使用求根公式求解

- 20 u^{2} - 36 u - 16 = 0

二次方程求根公式:

若

a = - 20, b = - 36, c = - 16

u_{1, 2} = \frac{- ( - 36 ) \pm ( - 36 ) ^{2} - 4 ( - 20 ) ( - 16 )}{2 ( - 20 )}

u_{1, 2} = \frac{- ( - 36 ) \pm ( - 36 ) ^{2} - 4 ( - 20 ) ( - 16 )}{2 ( - 20 )}

(- 36)^{2} - 4 (- 20) (- 16) = 4

(- 36)^{2} - 4 (- 20) (- 16)

使用法则

- (- a) = a

= (- 36)^{2} - 4 \cdot 20 \cdot 16

使用指数法则:

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

若

n

是偶数

(- 36)^{2} = 3 6^{2}

= 3 6^{2} - 4 \cdot 20 \cdot 16

数字相乘:

4 \cdot 20 \cdot 16 = 1280

= 3 6^{2} - 1280

3 6^{2} = 1296

= 1296 - 1280

数字相减:

1296 - 1280 = 16

= 16

因式分解数字:

16 = 4^{2}

= 4^{2}

使用根式运算法则:

n a^{n} = a

4^{2} = 4

= 4

u_{1, 2} = \frac{- ( - 36 ) \pm 4}{2 ( - 20 )}

将解分隔开

u_{1} = \frac{- ( - 36 ) + 4}{2 ( - 20 )}, u_{2} = \frac{- ( - 36 ) - 4}{2 ( - 20 )}

u = \frac{- ( - 36 ) + 4}{2 ( - 20 )} : - 1

\frac{- ( - 36 ) + 4}{2 ( - 20 )}

去除括号:

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

= \frac{36 + 4}{- 2 \cdot 20}

数字相加:

36 + 4 = 40

= \frac{40}{- 2 \cdot 20}

数字相乘:

2 \cdot 20 = 40

= \frac{40}{- 40}

使用分式法则:

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

= - \frac{40}{40}

使用法则

\frac{a}{a} = 1

= - 1

u = \frac{- ( - 36 ) - 4}{2 ( - 20 )} : - \frac{4}{5}

\frac{- ( - 36 ) - 4}{2 ( - 20 )}

去除括号:

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

= \frac{36 - 4}{- 2 \cdot 20}

数字相减:

36 - 4 = 32

= \frac{32}{- 2 \cdot 20}

数字相乘:

2 \cdot 20 = 40

= \frac{32}{- 40}

使用分式法则:

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

= - \frac{32}{40}

约分:

8

= - \frac{4}{5}

二次方程组的解是:

u = - 1, u = - \frac{4}{5}

u = cos (x)

代回

cos (x) = - 1, cos (x) = - \frac{4}{5}

cos (x) = - 1, cos (x) = - \frac{4}{5}

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{4}{5} : x = arccos (- \frac{4}{5}) + 2 πn, x = - arccos (- \frac{4}{5}) + 2 πn

cos (x) = - \frac{4}{5}

使用反三角函数性质

cos (x) = - \frac{4}{5}

cos (x) = - \frac{4}{5}

的通解

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

x = arccos (- \frac{4}{5}) + 2 πn, x = - arccos (- \frac{4}{5}) + 2 πn

x = arccos (- \frac{4}{5}) + 2 πn, x = - arccos (- \frac{4}{5}) + 2 πn

合并所有解

x = π + 2 πn, x = arccos (- \frac{4}{5}) + 2 πn, x = - arccos (- \frac{4}{5}) + 2 πn

以小数形式表示解

x = π + 2 πn, x = 2.49809 \dots + 2 πn, x = - 2.49809 \dots + 2 πn

作图

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