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受欢迎的三角函数 >

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

解答

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

解答

x = π + 2 πn, x = 1.28700 \dots + 2 πn

+1

度数

x = 18 0^{\circ} + 36 0^{\circ} n, x = 73.73979 \dots^{\circ} + 36 0^{\circ} n

求解步骤

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

两边加上

4 sin (x)

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

两边进行平方

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

两边减去

(- 3 + 4 sin (x))^{2}

9 cos^{2} (x) - 9 + 24 sin (x) - 16 sin^{2} (x) = 0

使用三角恒等式改写

- 9 - 16 sin^{2} (x) + 24 sin (x) + 9 cos^{2} (x)

使用毕达哥拉斯恒等式:

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

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

= - 9 - 16 sin^{2} (x) + 24 sin (x) + 9 (1 - sin^{2} (x))

化简

- 9 - 16 sin^{2} (x) + 24 sin (x) + 9 (1 - sin^{2} (x)) : 24 sin (x) - 25 sin^{2} (x)

- 9 - 16 sin^{2} (x) + 24 sin (x) + 9 (1 - sin^{2} (x))

乘开

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

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

使用分配律:

a (b - c) = ab - a c

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

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

数字相乘:

9 \cdot 1 = 9

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

= - 9 - 16 sin^{2} (x) + 24 sin (x) + 9 - 9 sin^{2} (x)

化简

- 9 - 16 sin^{2} (x) + 24 sin (x) + 9 - 9 sin^{2} (x) : 24 sin (x) - 25 sin^{2} (x)

- 9 - 16 sin^{2} (x) + 24 sin (x) + 9 - 9 sin^{2} (x)

对同类项分组

= - 16 sin^{2} (x) + 24 sin (x) - 9 sin^{2} (x) - 9 + 9

同类项相加:

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

= - 25 sin^{2} (x) + 24 sin (x) - 9 + 9

- 9 + 9 = 0

= 24 sin (x) - 25 sin^{2} (x)

= 24 sin (x) - 25 sin^{2} (x)

= 24 sin (x) - 25 sin^{2} (x)

24 sin (x) - 25 sin^{2} (x) = 0

用替代法求解

24 sin (x) - 25 sin^{2} (x) = 0

令

: sin (x) = u

24 u - 25 u^{2} = 0

24 u - 25 u^{2} = 0 : u = 0, u = \frac{24}{25}

24 u - 25 u^{2} = 0

改写成标准形式

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

- 25 u^{2} + 24 u = 0

使用求根公式求解

- 25 u^{2} + 24 u = 0

二次方程求根公式:

若

a = - 25, b = 24, c = 0

u_{1, 2} = \frac{- 24 \pm 2 4 ^{2} - 4 ( - 25 ) \cdot 0}{2 ( - 25 )}

u_{1, 2} = \frac{- 24 \pm 2 4 ^{2} - 4 ( - 25 ) \cdot 0}{2 ( - 25 )}

2 4^{2} - 4 (- 25) \cdot 0 = 24

2 4^{2} - 4 (- 25) \cdot 0

使用法则

- (- a) = a

= 2 4^{2} + 4 \cdot 25 \cdot 0

使用法则

0 \cdot a = 0

= 2 4^{2} + 0

2 4^{2} + 0 = 2 4^{2}

= 2 4^{2}

使用根式运算法则:

n a^{n} = a,

假定

a \geq 0

= 24

u_{1, 2} = \frac{- 24 \pm 24}{2 ( - 25 )}

将解分隔开

u_{1} = \frac{- 24 + 24}{2 ( - 25 )}, u_{2} = \frac{- 24 - 24}{2 ( - 25 )}

u = \frac{- 24 + 24}{2 ( - 25 )} : 0

\frac{- 24 + 24}{2 ( - 25 )}

去除括号:

(- a) = - a

= \frac{- 24 + 24}{- 2 \cdot 25}

数字相加/相减:

- 24 + 24 = 0

= \frac{0}{- 2 \cdot 25}

数字相乘:

2 \cdot 25 = 50

= \frac{0}{- 50}

使用分式法则:

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

= - \frac{0}{50}

使用法则

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

= - 0

= 0

u = \frac{- 24 - 24}{2 ( - 25 )} : \frac{24}{25}

\frac{- 24 - 24}{2 ( - 25 )}

去除括号:

(- a) = - a

= \frac{- 24 - 24}{- 2 \cdot 25}

数字相减:

- 24 - 24 = - 48

= \frac{- 48}{- 2 \cdot 25}

数字相乘:

2 \cdot 25 = 50

= \frac{- 48}{- 50}

使用分式法则:

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

= \frac{48}{50}

约分:

2

= \frac{24}{25}

二次方程组的解是:

u = 0, u = \frac{24}{25}

u = sin (x)

代回

sin (x) = 0, sin (x) = \frac{24}{25}

sin (x) = 0, sin (x) = \frac{24}{25}

sin (x) = 0 : x = 2 πn, x = π + 2 πn

sin (x) = 0

sin (x) = 0

的通解

sin (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} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{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} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

x = 0 + 2 πn, x = π + 2 πn

x = 0 + 2 πn, x = π + 2 πn

解

x = 0 + 2 πn : x = 2 πn

x = 0 + 2 πn

0 + 2 πn = 2 πn

x = 2 πn

x = 2 πn, x = π + 2 πn

sin (x) = \frac{24}{25} : x = arcsin (\frac{24}{25}) + 2 πn, x = π - arcsin (\frac{24}{25}) + 2 πn

sin (x) = \frac{24}{25}

使用反三角函数性质

sin (x) = \frac{24}{25}

sin (x) = \frac{24}{25}

的通解

sin (x) = a \Rightarrow x = arcsin (a) + 2 πn, x = π - arcsin (a) + 2 πn

x = arcsin (\frac{24}{25}) + 2 πn, x = π - arcsin (\frac{24}{25}) + 2 πn

x = arcsin (\frac{24}{25}) + 2 πn, x = π - arcsin (\frac{24}{25}) + 2 πn

合并所有解

x = 2 πn, x = π + 2 πn, x = arcsin (\frac{24}{25}) + 2 πn, x = π - arcsin (\frac{24}{25}) + 2 πn

将解代入原方程进行验证

将它们代入

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

检验解是否符合

去除与方程不符的解。

检验

2 πn

的解:

假

2 πn

代入

n = 1

2 π 1

对于

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

代入

x = 2 π 1

3 cos (2 π 1) - 4 sin (2 π 1) = - 3

整理后得

3 = - 3

\Rightarrow 假

检验

π + 2 πn

的解:

真

π + 2 πn

代入

n = 1

π + 2 π 1

对于

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

代入

x = π + 2 π 1

3 cos (π + 2 π 1) - 4 sin (π + 2 π 1) = - 3

整理后得

- 3 = - 3

\Rightarrow 真

检验

arcsin (\frac{24}{25}) + 2 πn

的解:

真

arcsin (\frac{24}{25}) + 2 πn

代入

n = 1

arcsin (\frac{24}{25}) + 2 π 1

对于

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

代入

x = arcsin (\frac{24}{25}) + 2 π 1

3 cos (arcsin (\frac{24}{25}) + 2 π 1) - 4 sin (arcsin (\frac{24}{25}) + 2 π 1) = - 3

整理后得

- 3 = - 3

\Rightarrow 真

检验

π - arcsin (\frac{24}{25}) + 2 πn

的解:

假

π - arcsin (\frac{24}{25}) + 2 πn

代入

n = 1

π - arcsin (\frac{24}{25}) + 2 π 1

对于

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

代入

x = π - arcsin (\frac{24}{25}) + 2 π 1

3 cos (π - arcsin (\frac{24}{25}) + 2 π 1) - 4 sin (π - arcsin (\frac{24}{25}) + 2 π 1) = - 3

整理后得

- 4.68 = - 3

\Rightarrow 假

x = π + 2 πn, x = arcsin (\frac{24}{25}) + 2 πn

以小数形式表示解

x = π + 2 πn, x = 1.28700 \dots + 2 πn

作图

流行的例子

6=50sin(x)-15cos(x),0<x< pi/2

6 = 50 sin (x) - 15 cos (x), 0 < x < \frac{π}{2}

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so l v e f or y, x = 3 sin (y)

sin (x) - 0.75 = 0

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

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tan(2x)+sec(2x)=4

tan (2 x) + sec (2 x) = 4