zs

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

受欢迎的三角函数 >

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

解答

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

解答

x = 0.64350 \dots + 2 πn

+1

度数

x = 36.86989 \dots^{\circ} + 36 0^{\circ} n

求解步骤

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

两边减去

3 sin (x)

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

两边进行平方

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

两边减去

(5 - 3 sin (x))^{2}

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

使用三角恒等式改写

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

化简

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

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

乘开

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

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

使用分配律:

a (b - c) = ab - a c

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

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

数字相乘:

16 \cdot 1 = 16

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

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

化简

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

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

对同类项分组

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

同类项相加:

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

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

数字相加/相减:

- 25 + 16 = - 9

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

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

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

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

用替代法求解

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

令

: sin (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 3 0 ^{2} - 4 ( - 25 ) ( - 9 )}{2 ( - 25 )}

u_{1, 2} = \frac{- 30 \pm 3 0 ^{2} - 4 ( - 25 ) ( - 9 )}{2 ( - 25 )}

3 0^{2} - 4 (- 25) (- 9) = 0

3 0^{2} - 4 (- 25) (- 9)

使用法则

- (- a) = a

= 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

= \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 = sin (x)

代回

sin (x) = \frac{3}{5}

sin (x) = \frac{3}{5}

sin (x) = \frac{3}{5} : x = arcsin (\frac{3}{5}) + 2 πn, x = π - arcsin (\frac{3}{5}) + 2 πn

sin (x) = \frac{3}{5}

使用反三角函数性质

sin (x) = \frac{3}{5}

sin (x) = \frac{3}{5}

的通解

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

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

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

合并所有解

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

将解代入原方程进行验证

将它们代入

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

检验解是否符合

去除与方程不符的解。

检验

arcsin (\frac{3}{5}) + 2 πn

的解:

真

arcsin (\frac{3}{5}) + 2 πn

代入

n = 1

arcsin (\frac{3}{5}) + 2 π 1

对于

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

代入

x = arcsin (\frac{3}{5}) + 2 π 1

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

整理后得

5 = 5

\Rightarrow 真

检验

π - arcsin (\frac{3}{5}) + 2 πn

的解:

假

π - arcsin (\frac{3}{5}) + 2 πn

代入

n = 1

π - arcsin (\frac{3}{5}) + 2 π 1

对于

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

代入

x = π - arcsin (\frac{3}{5}) + 2 π 1

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

整理后得

- 1.4 = 5

\Rightarrow 假

x = arcsin (\frac{3}{5}) + 2 πn

以小数形式表示解

x = 0.64350 \dots + 2 πn

作图

流行的例子

tan^2(θ)-sec^2(θ)=cos(-θ)

tan^{2} (θ) - sec^{2} (θ) = cos (- θ)

-2sin(t)-4cos(2t)=0

- 2 sin (t) - 4 cos (2 t) = 0

8sin(2x)-4sqrt(3)=0

8 sin (2 x) - 43 = 0

cos(3x)cos(9x)+sin(3x)sin(9x)=0

cos (3 x) cos (9 x) + sin (3 x) sin (9 x) = 0

cos(2x)+5sin(x)+2=0,0<= x<2pi

cos (2 x) + 5 sin (x) + 2 = 0, 0 \leq x < 2 π