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受欢迎的三角函数 >

20sin^2(x)+27cos(x)=6

解答

20 sin^{2} (x) + 27 cos (x) = 6

解答

x = 1.98231 \dots + 2 πn, x = - 1.98231 \dots + 2 πn

+1

度数

x = 113.57817 \dots^{\circ} + 36 0^{\circ} n, x = - 113.57817 \dots^{\circ} + 36 0^{\circ} n

求解步骤

20 sin^{2} (x) + 27 cos (x) = 6

两边减去

6

20 sin^{2} (x) + 27 cos (x) - 6 = 0

使用三角恒等式改写

- 6 + 20 sin^{2} (x) + 27 cos (x)

使用毕达哥拉斯恒等式:

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

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

= - 6 + 20 (1 - cos^{2} (x)) + 27 cos (x)

化简

- 6 + 20 (1 - cos^{2} (x)) + 27 cos (x) : 27 cos (x) - 20 cos^{2} (x) + 14

- 6 + 20 (1 - cos^{2} (x)) + 27 cos (x)

乘开

20 (1 - cos^{2} (x)) : 20 - 20 cos^{2} (x)

20 (1 - cos^{2} (x))

使用分配律:

a (b - c) = ab - a c

a = 20, b = 1, c = cos^{2} (x)

= 20 \cdot 1 - 20 cos^{2} (x)

数字相乘:

20 \cdot 1 = 20

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

= - 6 + 20 - 20 cos^{2} (x) + 27 cos (x)

数字相加/相减:

- 6 + 20 = 14

= 27 cos (x) - 20 cos^{2} (x) + 14

= 27 cos (x) - 20 cos^{2} (x) + 14

14 - 20 cos^{2} (x) + 27 cos (x) = 0

用替代法求解

14 - 20 cos^{2} (x) + 27 cos (x) = 0

令

: cos (x) = u

14 - 20 u^{2} + 27 u = 0

14 - 20 u^{2} + 27 u = 0 : u = - \frac{2}{5}, u = \frac{7}{4}

14 - 20 u^{2} + 27 u = 0

改写成标准形式

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

- 20 u^{2} + 27 u + 14 = 0

使用求根公式求解

- 20 u^{2} + 27 u + 14 = 0

二次方程求根公式:

若

a = - 20, b = 27, c = 14

u_{1, 2} = \frac{- 27 \pm 2 7 ^{2} - 4 ( - 20 ) \cdot 14}{2 ( - 20 )}

u_{1, 2} = \frac{- 27 \pm 2 7 ^{2} - 4 ( - 20 ) \cdot 14}{2 ( - 20 )}

2 7^{2} - 4 (- 20) \cdot 14 = 43

2 7^{2} - 4 (- 20) \cdot 14

使用法则

- (- a) = a

= 2 7^{2} + 4 \cdot 20 \cdot 14

数字相乘:

4 \cdot 20 \cdot 14 = 1120

= 2 7^{2} + 1120

2 7^{2} = 729

= 729 + 1120

数字相加:

729 + 1120 = 1849

= 1849

因式分解数字:

1849 = 4 3^{2}

= 4 3^{2}

使用根式运算法则:

n a^{n} = a

4 3^{2} = 43

= 43

u_{1, 2} = \frac{- 27 \pm 43}{2 ( - 20 )}

将解分隔开

u_{1} = \frac{- 27 + 43}{2 ( - 20 )}, u_{2} = \frac{- 27 - 43}{2 ( - 20 )}

u = \frac{- 27 + 43}{2 ( - 20 )} : - \frac{2}{5}

\frac{- 27 + 43}{2 ( - 20 )}

去除括号:

(- a) = - a

= \frac{- 27 + 43}{- 2 \cdot 20}

数字相加/相减:

- 27 + 43 = 16

= \frac{16}{- 2 \cdot 20}

数字相乘:

2 \cdot 20 = 40

= \frac{16}{- 40}

使用分式法则:

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

= - \frac{16}{40}

约分:

8

= - \frac{2}{5}

u = \frac{- 27 - 43}{2 ( - 20 )} : \frac{7}{4}

\frac{- 27 - 43}{2 ( - 20 )}

去除括号:

(- a) = - a

= \frac{- 27 - 43}{- 2 \cdot 20}

数字相减:

- 27 - 43 = - 70

= \frac{- 70}{- 2 \cdot 20}

数字相乘:

2 \cdot 20 = 40

= \frac{- 70}{- 40}

使用分式法则:

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

= \frac{70}{40}

约分:

10

= \frac{7}{4}

二次方程组的解是:

u = - \frac{2}{5}, u = \frac{7}{4}

u = cos (x)

代回

cos (x) = - \frac{2}{5}, cos (x) = \frac{7}{4}

cos (x) = - \frac{2}{5}, cos (x) = \frac{7}{4}

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

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

使用反三角函数性质

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

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

的通解

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

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

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

cos (x) = \frac{7}{4} :

无解

cos (x) = \frac{7}{4}

- 1 \leq cos (x) \leq 1

无解

合并所有解

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

以小数形式表示解

x = 1.98231 \dots + 2 πn, x = - 1.98231 \dots + 2 πn

作图

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