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受欢迎的三角函数 >

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

解答

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

解答

x = - 2.74680 \dots + 2 πn, x = \frac{π}{2} + 2 πn

+1

度数

x = - 157.38013 \dots^{\circ} + 36 0^{\circ} n, x = 9 0^{\circ} + 36 0^{\circ} n

求解步骤

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

两边加上

3 cos (x)

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

两边进行平方

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

两边减去

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

4 sin^{2} (x) - 4 - 12 cos (x) - 9 cos^{2} (x) = 0

使用三角恒等式改写

- 4 - 12 cos (x) + 4 sin^{2} (x) - 9 cos^{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 - 12 cos (x) + 4 (1 - cos^{2} (x)) - 9 cos^{2} (x)

化简

- 4 - 12 cos (x) + 4 (1 - cos^{2} (x)) - 9 cos^{2} (x) : - 13 cos^{2} (x) - 12 cos (x)

- 4 - 12 cos (x) + 4 (1 - cos^{2} (x)) - 9 cos^{2} (x)

乘开

4 (1 - cos^{2} (x)) : 4 - 4 cos^{2} (x)

4 (1 - cos^{2} (x))

使用分配律:

a (b - c) = ab - a c

a = 4, b = 1, c = cos^{2} (x)

= 4 \cdot 1 - 4 cos^{2} (x)

数字相乘:

4 \cdot 1 = 4

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

= - 4 - 12 cos (x) + 4 - 4 cos^{2} (x) - 9 cos^{2} (x)

化简

- 4 - 12 cos (x) + 4 - 4 cos^{2} (x) - 9 cos^{2} (x) : - 13 cos^{2} (x) - 12 cos (x)

- 4 - 12 cos (x) + 4 - 4 cos^{2} (x) - 9 cos^{2} (x)

同类项相加:

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

= - 4 - 12 cos (x) + 4 - 13 cos^{2} (x)

对同类项分组

= - 12 cos (x) - 13 cos^{2} (x) - 4 + 4

- 4 + 4 = 0

= - 13 cos^{2} (x) - 12 cos (x)

= - 13 cos^{2} (x) - 12 cos (x)

= - 13 cos^{2} (x) - 12 cos (x)

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

用替代法求解

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

令

: cos (x) = u

- 12 u - 13 u^{2} = 0

- 12 u - 13 u^{2} = 0 : u = - \frac{12}{13}, u = 0

- 12 u - 13 u^{2} = 0

改写成标准形式

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

- 13 u^{2} - 12 u = 0

使用求根公式求解

- 13 u^{2} - 12 u = 0

二次方程求根公式:

若

a = - 13, b = - 12, c = 0

u_{1, 2} = \frac{- ( - 12 ) \pm ( - 12 ) ^{2} - 4 ( - 13 ) \cdot 0}{2 ( - 13 )}

u_{1, 2} = \frac{- ( - 12 ) \pm ( - 12 ) ^{2} - 4 ( - 13 ) \cdot 0}{2 ( - 13 )}

(- 12)^{2} - 4 (- 13) \cdot 0 = 12

(- 12)^{2} - 4 (- 13) \cdot 0

使用法则

- (- a) = a

= (- 12)^{2} + 4 \cdot 13 \cdot 0

使用指数法则:

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

若

n

是偶数

(- 12)^{2} = 1 2^{2}

= 1 2^{2} + 4 \cdot 13 \cdot 0

使用法则

0 \cdot a = 0

= 1 2^{2} + 0

1 2^{2} + 0 = 1 2^{2}

= 1 2^{2}

使用根式运算法则:

n a^{n} = a,

假定

a \geq 0

= 12

u_{1, 2} = \frac{- ( - 12 ) \pm 12}{2 ( - 13 )}

将解分隔开

u_{1} = \frac{- ( - 12 ) + 12}{2 ( - 13 )}, u_{2} = \frac{- ( - 12 ) - 12}{2 ( - 13 )}

u = \frac{- ( - 12 ) + 12}{2 ( - 13 )} : - \frac{12}{13}

\frac{- ( - 12 ) + 12}{2 ( - 13 )}

去除括号:

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

= \frac{12 + 12}{- 2 \cdot 13}

数字相加:

12 + 12 = 24

= \frac{24}{- 2 \cdot 13}

数字相乘:

2 \cdot 13 = 26

= \frac{24}{- 26}

使用分式法则:

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

= - \frac{24}{26}

约分:

2

= - \frac{12}{13}

u = \frac{- ( - 12 ) - 12}{2 ( - 13 )} : 0

\frac{- ( - 12 ) - 12}{2 ( - 13 )}

去除括号:

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

= \frac{12 - 12}{- 2 \cdot 13}

数字相减:

12 - 12 = 0

= \frac{0}{- 2 \cdot 13}

数字相乘:

2 \cdot 13 = 26

= \frac{0}{- 26}

使用分式法则:

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

= - \frac{0}{26}

使用法则

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

= - 0

= 0

二次方程组的解是:

u = - \frac{12}{13}, u = 0

u = cos (x)

代回

cos (x) = - \frac{12}{13}, cos (x) = 0

cos (x) = - \frac{12}{13}, cos (x) = 0

cos (x) = - \frac{12}{13} : x = arccos (- \frac{12}{13}) + 2 πn, x = - arccos (- \frac{12}{13}) + 2 πn

cos (x) = - \frac{12}{13}

使用反三角函数性质

cos (x) = - \frac{12}{13}

cos (x) = - \frac{12}{13}

的通解

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

x = arccos (- \frac{12}{13}) + 2 πn, x = - arccos (- \frac{12}{13}) + 2 πn

x = arccos (- \frac{12}{13}) + 2 πn, x = - arccos (- \frac{12}{13}) + 2 πn

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

合并所有解

x = arccos (- \frac{12}{13}) + 2 πn, x = - arccos (- \frac{12}{13}) + 2 πn, x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

将解代入原方程进行验证

将它们代入

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

检验解是否符合

去除与方程不符的解。

检验

arccos (- \frac{12}{13}) + 2 πn

的解:

假

arccos (- \frac{12}{13}) + 2 πn

代入

n = 1

arccos (- \frac{12}{13}) + 2 π 1

对于

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

代入

x = arccos (- \frac{12}{13}) + 2 π 1

2 sin (arccos (- \frac{12}{13}) + 2 π 1) - 3 cos (arccos (- \frac{12}{13}) + 2 π 1) = 2

整理后得

3.53846 \dots = 2

\Rightarrow 假

检验

- arccos (- \frac{12}{13}) + 2 πn

的解:

真

- arccos (- \frac{12}{13}) + 2 πn

代入

n = 1

- arccos (- \frac{12}{13}) + 2 π 1

对于

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

代入

x = - arccos (- \frac{12}{13}) + 2 π 1

2 sin (- arccos (- \frac{12}{13}) + 2 π 1) - 3 cos (- arccos (- \frac{12}{13}) + 2 π 1) = 2

整理后得

2 = 2

\Rightarrow 真

检验

\frac{π}{2} + 2 πn

的解:

真

\frac{π}{2} + 2 πn

代入

n = 1

\frac{π}{2} + 2 π 1

对于

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

代入

x = \frac{π}{2} + 2 π 1

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

整理后得

2 = 2

\Rightarrow 真

检验

\frac{3 π}{2} + 2 πn

的解:

假

\frac{3 π}{2} + 2 πn

代入

n = 1

\frac{3 π}{2} + 2 π 1

对于

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

代入

x = \frac{3 π}{2} + 2 π 1

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

整理后得

- 2 = 2

\Rightarrow 假

x = - arccos (- \frac{12}{13}) + 2 πn, x = \frac{π}{2} + 2 πn

以小数形式表示解

x = - 2.74680 \dots + 2 πn, x = \frac{π}{2} + 2 πn

作图

流行的例子

sec (0)

sin (33 0^{\circ})

tan (θ) - 1 = 0

sin (θ) = - 1

cos (1)