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受欢迎的三角函数 >

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

解答

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

解答

x = - 0.92729 \dots + 2 πn

+1

度数

x = - 53.13010 \dots^{\circ} + 36 0^{\circ} n

求解步骤

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

两边加上

4 sin (x)

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

两边进行平方

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

两边减去

(5 + 4 sin (x))^{2}

9 cos^{2} (x) - 25 - 40 sin (x) - 16 sin^{2} (x) = 0

使用三角恒等式改写

- 25 - 16 sin^{2} (x) - 40 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)

= - 25 - 16 sin^{2} (x) - 40 sin (x) + 9 (1 - sin^{2} (x))

化简

- 25 - 16 sin^{2} (x) - 40 sin (x) + 9 (1 - sin^{2} (x)) : - 25 sin^{2} (x) - 40 sin (x) - 16

- 25 - 16 sin^{2} (x) - 40 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)

= - 25 - 16 sin^{2} (x) - 40 sin (x) + 9 - 9 sin^{2} (x)

化简

- 25 - 16 sin^{2} (x) - 40 sin (x) + 9 - 9 sin^{2} (x) : - 25 sin^{2} (x) - 40 sin (x) - 16

- 25 - 16 sin^{2} (x) - 40 sin (x) + 9 - 9 sin^{2} (x)

对同类项分组

= - 16 sin^{2} (x) - 40 sin (x) - 9 sin^{2} (x) - 25 + 9

同类项相加:

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

= - 25 sin^{2} (x) - 40 sin (x) - 25 + 9

数字相加/相减:

- 25 + 9 = - 16

= - 25 sin^{2} (x) - 40 sin (x) - 16

= - 25 sin^{2} (x) - 40 sin (x) - 16

= - 25 sin^{2} (x) - 40 sin (x) - 16

- 16 - 25 sin^{2} (x) - 40 sin (x) = 0

用替代法求解

- 16 - 25 sin^{2} (x) - 40 sin (x) = 0

令

: sin (x) = u

- 16 - 25 u^{2} - 40 u = 0

- 16 - 25 u^{2} - 40 u = 0 : u = - \frac{4}{5}

- 16 - 25 u^{2} - 40 u = 0

改写成标准形式

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

- 25 u^{2} - 40 u - 16 = 0

使用求根公式求解

- 25 u^{2} - 40 u - 16 = 0

二次方程求根公式:

若

a = - 25, b = - 40, c = - 16

u_{1, 2} = \frac{- ( - 40 ) \pm ( - 40 ) ^{2} - 4 ( - 25 ) ( - 16 )}{2 ( - 25 )}

u_{1, 2} = \frac{- ( - 40 ) \pm ( - 40 ) ^{2} - 4 ( - 25 ) ( - 16 )}{2 ( - 25 )}

(- 40)^{2} - 4 (- 25) (- 16) = 0

(- 40)^{2} - 4 (- 25) (- 16)

使用法则

- (- a) = a

= (- 40)^{2} - 4 \cdot 25 \cdot 16

使用指数法则:

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

若

n

是偶数

(- 40)^{2} = 4 0^{2}

= 4 0^{2} - 4 \cdot 25 \cdot 16

数字相乘:

4 \cdot 25 \cdot 16 = 1600

= 4 0^{2} - 1600

4 0^{2} = 1600

= 1600 - 1600

数字相减:

1600 - 1600 = 0

= 0

u_{1, 2} = \frac{- ( - 40 ) \pm 0}{2 ( - 25 )}

u = \frac{- ( - 40 )}{2 ( - 25 )}

\frac{- ( - 40 )}{2 ( - 25 )} = - \frac{4}{5}

\frac{- ( - 40 )}{2 ( - 25 )}

去除括号:

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

= \frac{40}{- 2 \cdot 25}

数字相乘:

2 \cdot 25 = 50

= \frac{40}{- 50}

使用分式法则:

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

= - \frac{40}{50}

约分:

10

= - \frac{4}{5}

u = - \frac{4}{5}

二次方程组的解是:

u = - \frac{4}{5}

u = sin (x)

代回

sin (x) = - \frac{4}{5}

sin (x) = - \frac{4}{5}

sin (x) = - \frac{4}{5} : x = arcsin (- \frac{4}{5}) + 2 πn, x = π + arcsin (\frac{4}{5}) + 2 πn

sin (x) = - \frac{4}{5}

使用反三角函数性质

sin (x) = - \frac{4}{5}

sin (x) = - \frac{4}{5}

的通解

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

x = arcsin (- \frac{4}{5}) + 2 πn, x = π + arcsin (\frac{4}{5}) + 2 πn

x = arcsin (- \frac{4}{5}) + 2 πn, x = π + arcsin (\frac{4}{5}) + 2 πn

合并所有解

x = arcsin (- \frac{4}{5}) + 2 πn, x = π + arcsin (\frac{4}{5}) + 2 πn

将解代入原方程进行验证

将它们代入

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

检验解是否符合

去除与方程不符的解。

检验

arcsin (- \frac{4}{5}) + 2 πn

的解:

真

arcsin (- \frac{4}{5}) + 2 πn

代入

n = 1

arcsin (- \frac{4}{5}) + 2 π 1

对于

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

代入

x = arcsin (- \frac{4}{5}) + 2 π 1

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

整理后得

5 = 5

\Rightarrow 真

检验

π + arcsin (\frac{4}{5}) + 2 πn

的解:

假

π + arcsin (\frac{4}{5}) + 2 πn

代入

n = 1

π + arcsin (\frac{4}{5}) + 2 π 1

对于

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

代入

x = π + arcsin (\frac{4}{5}) + 2 π 1

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

整理后得

1.4 = 5

\Rightarrow 假

x = arcsin (- \frac{4}{5}) + 2 πn

以小数形式表示解

x = - 0.92729 \dots + 2 πn

作图

流行的例子

sqrt(2)cos(x-pi/4)-1=0

2 cos (x - \frac{π}{4}) - 1 = 0

0 = 3 cos (x)

tan(pi/2-x)+tan(x)-1=0

tan (\frac{π}{2} - x) + tan (x) - 1 = 0

cos (1000 t) = 1

3sin(θ/4)-3/2 =0,0<= θ<= 2pi

3 sin (\frac{θ}{4}) - \frac{3}{2} = 0, 0 \leq θ \leq 2 π