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受欢迎的三角函数 >

4sin(x)-3cos(x)=5

解答

4 sin (x) - 3 cos (x) = 5

解答

x = 2.21429 \dots + 2 πn

+1

度数

x = 126.86989 \dots^{\circ} + 36 0^{\circ} n

求解步骤

4 sin (x) - 3 cos (x) = 5

两边加上

3 cos (x)

4 sin (x) = 5 + 3 cos (x)

两边进行平方

(4 sin (x))^{2} = (5 + 3 cos (x))^{2}

两边减去

(5 + 3 cos (x))^{2}

16 sin^{2} (x) - 25 - 30 cos (x) - 9 cos^{2} (x) = 0

使用三角恒等式改写

- 25 + 16 sin^{2} (x) - 30 cos (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)

= - 25 + 16 (1 - cos^{2} (x)) - 30 cos (x) - 9 cos^{2} (x)

化简

- 25 + 16 (1 - cos^{2} (x)) - 30 cos (x) - 9 cos^{2} (x) : - 25 cos^{2} (x) - 30 cos (x) - 9

- 25 + 16 (1 - cos^{2} (x)) - 30 cos (x) - 9 cos^{2} (x)

乘开

16 (1 - cos^{2} (x)) : 16 - 16 cos^{2} (x)

16 (1 - cos^{2} (x))

使用分配律:

a (b - c) = ab - a c

a = 16, b = 1, c = cos^{2} (x)

= 16 \cdot 1 - 16 cos^{2} (x)

数字相乘:

16 \cdot 1 = 16

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

= - 25 + 16 - 16 cos^{2} (x) - 30 cos (x) - 9 cos^{2} (x)

化简

- 25 + 16 - 16 cos^{2} (x) - 30 cos (x) - 9 cos^{2} (x) : - 25 cos^{2} (x) - 30 cos (x) - 9

- 25 + 16 - 16 cos^{2} (x) - 30 cos (x) - 9 cos^{2} (x)

对同类项分组

= - 16 cos^{2} (x) - 30 cos (x) - 9 cos^{2} (x) - 25 + 16

同类项相加:

- 16 cos^{2} (x) - 9 cos^{2} (x) = - 25 cos^{2} (x)

= - 25 cos^{2} (x) - 30 cos (x) - 25 + 16

数字相加/相减:

- 25 + 16 = - 9

= - 25 cos^{2} (x) - 30 cos (x) - 9

= - 25 cos^{2} (x) - 30 cos (x) - 9

= - 25 cos^{2} (x) - 30 cos (x) - 9

- 9 - 25 cos^{2} (x) - 30 cos (x) = 0

用替代法求解

- 9 - 25 cos^{2} (x) - 30 cos (x) = 0

令

: cos (x) = u

- 9 - 25 u^{2} - 30 u = 0

- 9 - 25 u^{2} - 30 u = 0 : u = - \frac{3}{5}

- 9 - 25 u^{2} - 30 u = 0

改写成标准形式

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

- 25 u^{2} - 30 u - 9 = 0

使用求根公式求解

- 25 u^{2} - 30 u - 9 = 0

二次方程求根公式:

若

a = - 25, b = - 30, c = - 9

u_{1, 2} = \frac{- ( - 30 ) \pm ( - 30 ) ^{2} - 4 ( - 25 ) ( - 9 )}{2 ( - 25 )}

u_{1, 2} = \frac{- ( - 30 ) \pm ( - 30 ) ^{2} - 4 ( - 25 ) ( - 9 )}{2 ( - 25 )}

(- 30)^{2} - 4 (- 25) (- 9) = 0

(- 30)^{2} - 4 (- 25) (- 9)

使用法则

- (- a) = a

= (- 30)^{2} - 4 \cdot 25 \cdot 9

使用指数法则:

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

若

n

是偶数

(- 30)^{2} = 3 0^{2}

= 3 0^{2} - 4 \cdot 25 \cdot 9

数字相乘:

4 \cdot 25 \cdot 9 = 900

= 3 0^{2} - 900

3 0^{2} = 900

= 900 - 900

数字相减:

900 - 900 = 0

= 0

u_{1, 2} = \frac{- ( - 30 ) \pm 0}{2 ( - 25 )}

u = \frac{- ( - 30 )}{2 ( - 25 )}

\frac{- ( - 30 )}{2 ( - 25 )} = - \frac{3}{5}

\frac{- ( - 30 )}{2 ( - 25 )}

去除括号:

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

= \frac{30}{- 2 \cdot 25}

数字相乘:

2 \cdot 25 = 50

= \frac{30}{- 50}

使用分式法则:

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

= - \frac{30}{50}

约分:

10

= - \frac{3}{5}

u = - \frac{3}{5}

二次方程组的解是:

u = - \frac{3}{5}

u = cos (x)

代回

cos (x) = - \frac{3}{5}

cos (x) = - \frac{3}{5}

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

cos (x) = - \frac{3}{5}

使用反三角函数性质

cos (x) = - \frac{3}{5}

cos (x) = - \frac{3}{5}

的通解

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

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

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

合并所有解

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

将解代入原方程进行验证

将它们代入

4 sin (x) - 3 cos (x) = 5

检验解是否符合

去除与方程不符的解。

检验

arccos (- \frac{3}{5}) + 2 πn

的解:

真

arccos (- \frac{3}{5}) + 2 πn

代入

n = 1

arccos (- \frac{3}{5}) + 2 π 1

对于

4 sin (x) - 3 cos (x) = 5

代入

x = arccos (- \frac{3}{5}) + 2 π 1

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

整理后得

5 = 5

\Rightarrow 真

检验

- arccos (- \frac{3}{5}) + 2 πn

的解:

假

- arccos (- \frac{3}{5}) + 2 πn

代入

n = 1

- arccos (- \frac{3}{5}) + 2 π 1

对于

4 sin (x) - 3 cos (x) = 5

代入

x = - arccos (- \frac{3}{5}) + 2 π 1

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

整理后得

- 1.4 = 5

\Rightarrow 假

x = arccos (- \frac{3}{5}) + 2 πn

以小数形式表示解

x = 2.21429 \dots + 2 πn

作图

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