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受欢迎的三角函数 >

-2cos(x)+3sin(x)=2

解答

- 2 cos (x) + 3 sin (x) = 2

解答

x = π + 2 πn, x = 1.17600 \dots + 2 πn

+1

度数

x = 18 0^{\circ} + 36 0^{\circ} n, x = 67.38013 \dots^{\circ} + 36 0^{\circ} n

求解步骤

- 2 cos (x) + 3 sin (x) = 2

两边加上

2 cos (x)

3 sin (x) = 2 + 2 cos (x)

两边进行平方

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

两边减去

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

9 sin^{2} (x) - 4 - 8 cos (x) - 4 cos^{2} (x) = 0

使用三角恒等式改写

- 4 - 4 cos^{2} (x) - 8 cos (x) + 9 sin^{2} (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)

= - 4 - 4 cos^{2} (x) - 8 cos (x) + 9 (1 - cos^{2} (x))

化简

- 4 - 4 cos^{2} (x) - 8 cos (x) + 9 (1 - cos^{2} (x)) : - 13 cos^{2} (x) - 8 cos (x) + 5

- 4 - 4 cos^{2} (x) - 8 cos (x) + 9 (1 - cos^{2} (x))

乘开

9 (1 - cos^{2} (x)) : 9 - 9 cos^{2} (x)

9 (1 - cos^{2} (x))

使用分配律:

a (b - c) = ab - a c

a = 9, b = 1, c = cos^{2} (x)

= 9 \cdot 1 - 9 cos^{2} (x)

数字相乘:

9 \cdot 1 = 9

= 9 - 9 cos^{2} (x)

= - 4 - 4 cos^{2} (x) - 8 cos (x) + 9 - 9 cos^{2} (x)

化简

- 4 - 4 cos^{2} (x) - 8 cos (x) + 9 - 9 cos^{2} (x) : - 13 cos^{2} (x) - 8 cos (x) + 5

- 4 - 4 cos^{2} (x) - 8 cos (x) + 9 - 9 cos^{2} (x)

对同类项分组

= - 4 cos^{2} (x) - 8 cos (x) - 9 cos^{2} (x) - 4 + 9

同类项相加:

- 4 cos^{2} (x) - 9 cos^{2} (x) = - 13 cos^{2} (x)

= - 13 cos^{2} (x) - 8 cos (x) - 4 + 9

数字相加/相减:

- 4 + 9 = 5

= - 13 cos^{2} (x) - 8 cos (x) + 5

= - 13 cos^{2} (x) - 8 cos (x) + 5

= - 13 cos^{2} (x) - 8 cos (x) + 5

5 - 13 cos^{2} (x) - 8 cos (x) = 0

用替代法求解

5 - 13 cos^{2} (x) - 8 cos (x) = 0

令

: cos (x) = u

5 - 13 u^{2} - 8 u = 0

5 - 13 u^{2} - 8 u = 0 : u = - 1, u = \frac{5}{13}

5 - 13 u^{2} - 8 u = 0

改写成标准形式

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

- 13 u^{2} - 8 u + 5 = 0

使用求根公式求解

- 13 u^{2} - 8 u + 5 = 0

二次方程求根公式:

若

a = - 13, b = - 8, c = 5

u_{1, 2} = \frac{- ( - 8 ) \pm ( - 8 ) ^{2} - 4 ( - 13 ) \cdot 5}{2 ( - 13 )}

u_{1, 2} = \frac{- ( - 8 ) \pm ( - 8 ) ^{2} - 4 ( - 13 ) \cdot 5}{2 ( - 13 )}

(- 8)^{2} - 4 (- 13) \cdot 5 = 18

(- 8)^{2} - 4 (- 13) \cdot 5

使用法则

- (- a) = a

= (- 8)^{2} + 4 \cdot 13 \cdot 5

使用指数法则:

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

若

n

是偶数

(- 8)^{2} = 8^{2}

= 8^{2} + 4 \cdot 13 \cdot 5

数字相乘:

4 \cdot 13 \cdot 5 = 260

= 8^{2} + 260

8^{2} = 64

= 64 + 260

数字相加:

64 + 260 = 324

= 324

因式分解数字:

324 = 1 8^{2}

= 1 8^{2}

使用根式运算法则:

n a^{n} = a

1 8^{2} = 18

= 18

u_{1, 2} = \frac{- ( - 8 ) \pm 18}{2 ( - 13 )}

将解分隔开

u_{1} = \frac{- ( - 8 ) + 18}{2 ( - 13 )}, u_{2} = \frac{- ( - 8 ) - 18}{2 ( - 13 )}

u = \frac{- ( - 8 ) + 18}{2 ( - 13 )} : - 1

\frac{- ( - 8 ) + 18}{2 ( - 13 )}

去除括号:

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

= \frac{8 + 18}{- 2 \cdot 13}

数字相加:

8 + 18 = 26

= \frac{26}{- 2 \cdot 13}

数字相乘:

2 \cdot 13 = 26

= \frac{26}{- 26}

使用分式法则:

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

= - \frac{26}{26}

使用法则

\frac{a}{a} = 1

= - 1

u = \frac{- ( - 8 ) - 18}{2 ( - 13 )} : \frac{5}{13}

\frac{- ( - 8 ) - 18}{2 ( - 13 )}

去除括号:

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

= \frac{8 - 18}{- 2 \cdot 13}

数字相减:

8 - 18 = - 10

= \frac{- 10}{- 2 \cdot 13}

数字相乘:

2 \cdot 13 = 26

= \frac{- 10}{- 26}

使用分式法则:

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

= \frac{10}{26}

约分:

2

= \frac{5}{13}

二次方程组的解是:

u = - 1, u = \frac{5}{13}

u = cos (x)

代回

cos (x) = - 1, cos (x) = \frac{5}{13}

cos (x) = - 1, cos (x) = \frac{5}{13}

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{5}{13} : x = arccos (\frac{5}{13}) + 2 πn, x = 2 π - arccos (\frac{5}{13}) + 2 πn

cos (x) = \frac{5}{13}

使用反三角函数性质

cos (x) = \frac{5}{13}

cos (x) = \frac{5}{13}

的通解

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

x = arccos (\frac{5}{13}) + 2 πn, x = 2 π - arccos (\frac{5}{13}) + 2 πn

x = arccos (\frac{5}{13}) + 2 πn, x = 2 π - arccos (\frac{5}{13}) + 2 πn

合并所有解

x = π + 2 πn, x = arccos (\frac{5}{13}) + 2 πn, x = 2 π - arccos (\frac{5}{13}) + 2 πn

将解代入原方程进行验证

将它们代入

- 2 cos (x) + 3 sin (x) = 2

检验解是否符合

去除与方程不符的解。

检验

π + 2 πn

的解:

真

π + 2 πn

代入

n = 1

π + 2 π 1

对于

- 2 cos (x) + 3 sin (x) = 2

代入

x = π + 2 π 1

- 2 cos (π + 2 π 1) + 3 sin (π + 2 π 1) = 2

整理后得

2 = 2

\Rightarrow 真

检验

arccos (\frac{5}{13}) + 2 πn

的解:

真

arccos (\frac{5}{13}) + 2 πn

代入

n = 1

arccos (\frac{5}{13}) + 2 π 1

对于

- 2 cos (x) + 3 sin (x) = 2

代入

x = arccos (\frac{5}{13}) + 2 π 1

- 2 cos (arccos (\frac{5}{13}) + 2 π 1) + 3 sin (arccos (\frac{5}{13}) + 2 π 1) = 2

整理后得

2 = 2

\Rightarrow 真

检验

2 π - arccos (\frac{5}{13}) + 2 πn

的解:

假

2 π - arccos (\frac{5}{13}) + 2 πn

代入

n = 1

2 π - arccos (\frac{5}{13}) + 2 π 1

对于

- 2 cos (x) + 3 sin (x) = 2

代入

x = 2 π - arccos (\frac{5}{13}) + 2 π 1

- 2 cos (2 π - arccos (\frac{5}{13}) + 2 π 1) + 3 sin (2 π - arccos (\frac{5}{13}) + 2 π 1) = 2

整理后得

- 3.53846 \dots = 2

\Rightarrow 假

x = π + 2 πn, x = arccos (\frac{5}{13}) + 2 πn

以小数形式表示解

x = π + 2 πn, x = 1.17600 \dots + 2 πn

作图

流行的例子

cos(x+pi/6)-1=cos(x-pi/6)

cos (x + \frac{π}{6}) - 1 = cos (x - \frac{π}{6})

sin (x) = \frac{2}{7}

6cos^2(x)-cos(x)-1=0

6 cos^{2} (x) - cos (x) - 1 = 0

tan (x) = 0.839

sqrt(3)-tan(3θ)=0

3 - tan (3 θ) = 0