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 >

5cos(x)+2sin(x)=2

解答

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

解答

x = - 0.80978 \dots + 2 πn, x = \frac{π}{2} + 2 πn

+1

度数

x = - 46.39718 \dots^{\circ} + 36 0^{\circ} n, x = 9 0^{\circ} + 36 0^{\circ} n

求解步骤

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

两边减去

2 sin (x)

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

两边进行平方

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

两边减去

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

25 cos^{2} (x) - 4 + 8 sin (x) - 4 sin^{2} (x) = 0

使用三角恒等式改写

- 4 + 25 cos^{2} (x) - 4 sin^{2} (x) + 8 sin (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)

= - 4 + 25 (1 - sin^{2} (x)) - 4 sin^{2} (x) + 8 sin (x)

化简

- 4 + 25 (1 - sin^{2} (x)) - 4 sin^{2} (x) + 8 sin (x) : 8 sin (x) - 29 sin^{2} (x) + 21

- 4 + 25 (1 - sin^{2} (x)) - 4 sin^{2} (x) + 8 sin (x)

乘开

25 (1 - sin^{2} (x)) : 25 - 25 sin^{2} (x)

25 (1 - sin^{2} (x))

使用分配律:

a (b - c) = ab - a c

a = 25, b = 1, c = sin^{2} (x)

= 25 \cdot 1 - 25 sin^{2} (x)

数字相乘:

25 \cdot 1 = 25

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

= - 4 + 25 - 25 sin^{2} (x) - 4 sin^{2} (x) + 8 sin (x)

化简

- 4 + 25 - 25 sin^{2} (x) - 4 sin^{2} (x) + 8 sin (x) : 8 sin (x) - 29 sin^{2} (x) + 21

- 4 + 25 - 25 sin^{2} (x) - 4 sin^{2} (x) + 8 sin (x)

同类项相加:

- 25 sin^{2} (x) - 4 sin^{2} (x) = - 29 sin^{2} (x)

= - 4 + 25 - 29 sin^{2} (x) + 8 sin (x)

数字相加/相减:

- 4 + 25 = 21

= 8 sin (x) - 29 sin^{2} (x) + 21

= 8 sin (x) - 29 sin^{2} (x) + 21

= 8 sin (x) - 29 sin^{2} (x) + 21

21 - 29 sin^{2} (x) + 8 sin (x) = 0

用替代法求解

21 - 29 sin^{2} (x) + 8 sin (x) = 0

令

: sin (x) = u

21 - 29 u^{2} + 8 u = 0

21 - 29 u^{2} + 8 u = 0 : u = - \frac{21}{29}, u = 1

21 - 29 u^{2} + 8 u = 0

改写成标准形式

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

- 29 u^{2} + 8 u + 21 = 0

使用求根公式求解

- 29 u^{2} + 8 u + 21 = 0

二次方程求根公式:

若

a = - 29, b = 8, c = 21

u_{1, 2} = \frac{- 8 \pm 8 ^{2} - 4 ( - 29 ) \cdot 21}{2 ( - 29 )}

u_{1, 2} = \frac{- 8 \pm 8 ^{2} - 4 ( - 29 ) \cdot 21}{2 ( - 29 )}

8^{2} - 4 (- 29) \cdot 21 = 50

8^{2} - 4 (- 29) \cdot 21

使用法则

- (- a) = a

= 8^{2} + 4 \cdot 29 \cdot 21

数字相乘:

4 \cdot 29 \cdot 21 = 2436

= 8^{2} + 2436

8^{2} = 64

= 64 + 2436

数字相加:

64 + 2436 = 2500

= 2500

因式分解数字:

2500 = 5 0^{2}

= 5 0^{2}

使用根式运算法则:

n a^{n} = a

5 0^{2} = 50

= 50

u_{1, 2} = \frac{- 8 \pm 50}{2 ( - 29 )}

将解分隔开

u_{1} = \frac{- 8 + 50}{2 ( - 29 )}, u_{2} = \frac{- 8 - 50}{2 ( - 29 )}

u = \frac{- 8 + 50}{2 ( - 29 )} : - \frac{21}{29}

\frac{- 8 + 50}{2 ( - 29 )}

去除括号:

(- a) = - a

= \frac{- 8 + 50}{- 2 \cdot 29}

数字相加/相减:

- 8 + 50 = 42

= \frac{42}{- 2 \cdot 29}

数字相乘:

2 \cdot 29 = 58

= \frac{42}{- 58}

使用分式法则:

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

= - \frac{42}{58}

约分:

2

= - \frac{21}{29}

u = \frac{- 8 - 50}{2 ( - 29 )} : 1

\frac{- 8 - 50}{2 ( - 29 )}

去除括号:

(- a) = - a

= \frac{- 8 - 50}{- 2 \cdot 29}

数字相减:

- 8 - 50 = - 58

= \frac{- 58}{- 2 \cdot 29}

数字相乘:

2 \cdot 29 = 58

= \frac{- 58}{- 58}

使用分式法则:

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

= \frac{58}{58}

使用法则

\frac{a}{a} = 1

= 1

二次方程组的解是:

u = - \frac{21}{29}, u = 1

u = sin (x)

代回

sin (x) = - \frac{21}{29}, sin (x) = 1

sin (x) = - \frac{21}{29}, sin (x) = 1

sin (x) = - \frac{21}{29} : x = arcsin (- \frac{21}{29}) + 2 πn, x = π + arcsin (\frac{21}{29}) + 2 πn

sin (x) = - \frac{21}{29}

使用反三角函数性质

sin (x) = - \frac{21}{29}

sin (x) = - \frac{21}{29}

的通解

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

x = arcsin (- \frac{21}{29}) + 2 πn, x = π + arcsin (\frac{21}{29}) + 2 πn

x = arcsin (- \frac{21}{29}) + 2 πn, x = π + arcsin (\frac{21}{29}) + 2 πn

sin (x) = 1 : x = \frac{π}{2} + 2 πn

sin (x) = 1

sin (x) = 1

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 = \frac{π}{2} + 2 πn

x = \frac{π}{2} + 2 πn

合并所有解

x = arcsin (- \frac{21}{29}) + 2 πn, x = π + arcsin (\frac{21}{29}) + 2 πn, x = \frac{π}{2} + 2 πn

将解代入原方程进行验证

将它们代入

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

检验解是否符合

去除与方程不符的解。

检验

arcsin (- \frac{21}{29}) + 2 πn

的解:

真

arcsin (- \frac{21}{29}) + 2 πn

代入

n = 1

arcsin (- \frac{21}{29}) + 2 π 1

对于

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

代入

x = arcsin (- \frac{21}{29}) + 2 π 1

5 cos (arcsin (- \frac{21}{29}) + 2 π 1) + 2 sin (arcsin (- \frac{21}{29}) + 2 π 1) = 2

整理后得

2 = 2

\Rightarrow 真

检验

π + arcsin (\frac{21}{29}) + 2 πn

的解:

假

π + arcsin (\frac{21}{29}) + 2 πn

代入

n = 1

π + arcsin (\frac{21}{29}) + 2 π 1

对于

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

代入

x = π + arcsin (\frac{21}{29}) + 2 π 1

5 cos (π + arcsin (\frac{21}{29}) + 2 π 1) + 2 sin (π + arcsin (\frac{21}{29}) + 2 π 1) = 2

整理后得

- 4.89655 \dots = 2

\Rightarrow 假

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

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

代入

x = \frac{π}{2} + 2 π 1

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

整理后得

2 = 2

\Rightarrow 真

x = arcsin (- \frac{21}{29}) + 2 πn, x = \frac{π}{2} + 2 πn

以小数形式表示解

x = - 0.80978 \dots + 2 πn, x = \frac{π}{2} + 2 πn

作图

流行的例子

sin(2θ)+sin(θ)+2cos(θ)+1=0,0<= θ<= 2pi

sin (2 θ) + sin (θ) + 2 cos (θ) + 1 = 0, 0 \leq θ \leq 2 π

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

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

3sec^2(θ)-12=5sec(θ)

3 sec^{2} (θ) - 12 = 5 sec (θ)

3cos(x)-sqrt(3)=cos(x)

3 cos (x) - 3 = cos (x)

36 cos (2 x) = 0