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受欢迎的三角函数 >

3cos(x)=2-sin(x)

解答

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

解答

x = - 0.56432 \dots + 2 πn, x = 1.20782 \dots + 2 πn

+1

度数

x = - 32.33353 \dots^{\circ} + 36 0^{\circ} n, x = 69.20342 \dots^{\circ} + 36 0^{\circ} n

求解步骤

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

两边进行平方

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

两边减去

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

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

使用三角恒等式改写

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

= - 4 - sin^{2} (x) + 4 sin (x) + 9 (1 - sin^{2} (x))

化简

- 4 - sin^{2} (x) + 4 sin (x) + 9 (1 - sin^{2} (x)) : 4 sin (x) - 10 sin^{2} (x) + 5

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

= - 4 - sin^{2} (x) + 4 sin (x) + 9 - 9 sin^{2} (x)

化简

- 4 - sin^{2} (x) + 4 sin (x) + 9 - 9 sin^{2} (x) : 4 sin (x) - 10 sin^{2} (x) + 5

- 4 - sin^{2} (x) + 4 sin (x) + 9 - 9 sin^{2} (x)

对同类项分组

= - sin^{2} (x) + 4 sin (x) - 9 sin^{2} (x) - 4 + 9

同类项相加:

- sin^{2} (x) - 9 sin^{2} (x) = - 10 sin^{2} (x)

= - 10 sin^{2} (x) + 4 sin (x) - 4 + 9

数字相加/相减:

- 4 + 9 = 5

= 4 sin (x) - 10 sin^{2} (x) + 5

= 4 sin (x) - 10 sin^{2} (x) + 5

= 4 sin (x) - 10 sin^{2} (x) + 5

5 - 10 sin^{2} (x) + 4 sin (x) = 0

用替代法求解

5 - 10 sin^{2} (x) + 4 sin (x) = 0

令

: sin (x) = u

5 - 10 u^{2} + 4 u = 0

5 - 10 u^{2} + 4 u = 0 : u = - \frac{- 2 + 3 6}{10}, u = \frac{2 + 3 6}{10}

5 - 10 u^{2} + 4 u = 0

改写成标准形式

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

- 10 u^{2} + 4 u + 5 = 0

使用求根公式求解

- 10 u^{2} + 4 u + 5 = 0

二次方程求根公式:

若

a = - 10, b = 4, c = 5

u_{1, 2} = \frac{- 4 \pm 4 ^{2} - 4 ( - 10 ) \cdot 5}{2 ( - 10 )}

u_{1, 2} = \frac{- 4 \pm 4 ^{2} - 4 ( - 10 ) \cdot 5}{2 ( - 10 )}

4^{2} - 4 (- 10) \cdot 5 = 66

4^{2} - 4 (- 10) \cdot 5

使用法则

- (- a) = a

= 4^{2} + 4 \cdot 10 \cdot 5

数字相乘:

4 \cdot 10 \cdot 5 = 200

= 4^{2} + 200

4^{2} = 16

= 16 + 200

数字相加:

16 + 200 = 216

= 216

216

质因数分解:

2^{3} \cdot 3^{3}

216

216

除以

2216 = 108 \cdot 2

= 2 \cdot 108

108

除以

2108 = 54 \cdot 2

= 2 \cdot 2 \cdot 54

54

除以

254 = 27 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 27

27

除以

327 = 9 \cdot 3

= 2 \cdot 2 \cdot 2 \cdot 3 \cdot 9

9

除以

39 = 3 \cdot 3

= 2 \cdot 2 \cdot 2 \cdot 3 \cdot 3 \cdot 3

2, 3

都是质数，因此无法进一步因数分解

= 2 \cdot 2 \cdot 2 \cdot 3 \cdot 3 \cdot 3

= 2^{3} \cdot 3^{3}

= 2^{3} \cdot 3^{3}

使用指数法则:

a^{b + c} = a^{b} \cdot a^{c}

= 2^{2} \cdot 3^{2} \cdot 2 \cdot 3

使用根式运算法则:

n ab = n a n b

= 2^{2} 3^{2} 2 \cdot 3

使用根式运算法则:

n a^{n} = a

2^{2} = 2

= 2 3^{2} 2 \cdot 3

使用根式运算法则:

n a^{n} = a

3^{2} = 3

= 2 \cdot 3 2 \cdot 3

整理后得

= 66

u_{1, 2} = \frac{- 4 \pm 6 6}{2 ( - 10 )}

将解分隔开

u_{1} = \frac{- 4 + 6 6}{2 ( - 10 )}, u_{2} = \frac{- 4 - 6 6}{2 ( - 10 )}

u = \frac{- 4 + 6 6}{2 ( - 10 )} : - \frac{- 2 + 3 6}{10}

\frac{- 4 + 6 6}{2 ( - 10 )}

去除括号:

(- a) = - a

= \frac{- 4 + 6 6}{- 2 \cdot 10}

数字相乘:

2 \cdot 10 = 20

= \frac{- 4 + 6 6}{- 20}

使用分式法则:

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

= - \frac{- 4 + 6 6}{20}

消掉

\frac{- 4 + 6 6}{20} : \frac{3 6 - 2}{10}

\frac{- 4 + 6 6}{20}

分解

- 4 + 66 : 2 (- 2 + 36)

- 4 + 66

改写为

= - 2 \cdot 2 + 2 \cdot 36

因式分解出通项

2

= 2 (- 2 + 36)

= \frac{2 ( - 2 + 3 6 )}{20}

约分:

2

= \frac{- 2 + 3 6}{10}

= - \frac{3 6 - 2}{10}

= - \frac{- 2 + 3 6}{10}

u = \frac{- 4 - 6 6}{2 ( - 10 )} : \frac{2 + 3 6}{10}

\frac{- 4 - 6 6}{2 ( - 10 )}

去除括号:

(- a) = - a

= \frac{- 4 - 6 6}{- 2 \cdot 10}

数字相乘:

2 \cdot 10 = 20

= \frac{- 4 - 6 6}{- 20}

使用分式法则:

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

- 4 - 66 = - (4 + 66)

= \frac{4 + 6 6}{20}

分解

4 + 66 : 2 (2 + 36)

4 + 66

改写为

= 2 \cdot 2 + 2 \cdot 36

因式分解出通项

2

= 2 (2 + 36)

= \frac{2 ( 2 + 3 6 )}{20}

约分:

2

= \frac{2 + 3 6}{10}

二次方程组的解是:

u = - \frac{- 2 + 3 6}{10}, u = \frac{2 + 3 6}{10}

u = sin (x)

代回

sin (x) = - \frac{- 2 + 3 6}{10}, sin (x) = \frac{2 + 3 6}{10}

sin (x) = - \frac{- 2 + 3 6}{10}, sin (x) = \frac{2 + 3 6}{10}

sin (x) = - \frac{- 2 + 3 6}{10} : x = arcsin (- \frac{- 2 + 3 6}{10}) + 2 πn, x = π + arcsin (\frac{- 2 + 3 6}{10}) + 2 πn

sin (x) = - \frac{- 2 + 3 6}{10}

使用反三角函数性质

sin (x) = - \frac{- 2 + 3 6}{10}

sin (x) = - \frac{- 2 + 3 6}{10}

的通解

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

x = arcsin (- \frac{- 2 + 3 6}{10}) + 2 πn, x = π + arcsin (\frac{- 2 + 3 6}{10}) + 2 πn

x = arcsin (- \frac{- 2 + 3 6}{10}) + 2 πn, x = π + arcsin (\frac{- 2 + 3 6}{10}) + 2 πn

sin (x) = \frac{2 + 3 6}{10} : x = arcsin (\frac{2 + 3 6}{10}) + 2 πn, x = π - arcsin (\frac{2 + 3 6}{10}) + 2 πn

sin (x) = \frac{2 + 3 6}{10}

使用反三角函数性质

sin (x) = \frac{2 + 3 6}{10}

sin (x) = \frac{2 + 3 6}{10}

的通解

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

x = arcsin (\frac{2 + 3 6}{10}) + 2 πn, x = π - arcsin (\frac{2 + 3 6}{10}) + 2 πn

x = arcsin (\frac{2 + 3 6}{10}) + 2 πn, x = π - arcsin (\frac{2 + 3 6}{10}) + 2 πn

合并所有解

x = arcsin (- \frac{- 2 + 3 6}{10}) + 2 πn, x = π + arcsin (\frac{- 2 + 3 6}{10}) + 2 πn, x = arcsin (\frac{2 + 3 6}{10}) + 2 πn, x = π - arcsin (\frac{2 + 3 6}{10}) + 2 πn

将解代入原方程进行验证

将它们代入

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

检验解是否符合

去除与方程不符的解。

检验

arcsin (- \frac{- 2 + 3 6}{10}) + 2 πn

的解:

真

arcsin (- \frac{- 2 + 3 6}{10}) + 2 πn

代入

n = 1

arcsin (- \frac{- 2 + 3 6}{10}) + 2 π 1

对于

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

代入

x = arcsin (- \frac{- 2 + 3 6}{10}) + 2 π 1

3 cos (arcsin (- \frac{- 2 + 3 6}{10}) + 2 π 1) = 2 - sin (arcsin (- \frac{- 2 + 3 6}{10}) + 2 π 1)

整理后得

2.53484 \dots = 2.53484 \dots

\Rightarrow 真

检验

π + arcsin (\frac{- 2 + 3 6}{10}) + 2 πn

的解:

假

π + arcsin (\frac{- 2 + 3 6}{10}) + 2 πn

代入

n = 1

π + arcsin (\frac{- 2 + 3 6}{10}) + 2 π 1

对于

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

代入

x = π + arcsin (\frac{- 2 + 3 6}{10}) + 2 π 1

3 cos (π + arcsin (\frac{- 2 + 3 6}{10}) + 2 π 1) = 2 - sin (π + arcsin (\frac{- 2 + 3 6}{10}) + 2 π 1)

整理后得

- 2.53484 \dots = 2.53484 \dots

\Rightarrow 假

检验

arcsin (\frac{2 + 3 6}{10}) + 2 πn

的解:

真

arcsin (\frac{2 + 3 6}{10}) + 2 πn

代入

n = 1

arcsin (\frac{2 + 3 6}{10}) + 2 π 1

对于

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

代入

x = arcsin (\frac{2 + 3 6}{10}) + 2 π 1

3 cos (arcsin (\frac{2 + 3 6}{10}) + 2 π 1) = 2 - sin (arcsin (\frac{2 + 3 6}{10}) + 2 π 1)

整理后得

1.06515 \dots = 1.06515 \dots

\Rightarrow 真

检验

π - arcsin (\frac{2 + 3 6}{10}) + 2 πn

的解:

假

π - arcsin (\frac{2 + 3 6}{10}) + 2 πn

代入

n = 1

π - arcsin (\frac{2 + 3 6}{10}) + 2 π 1

对于

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

代入

x = π - arcsin (\frac{2 + 3 6}{10}) + 2 π 1

3 cos (π - arcsin (\frac{2 + 3 6}{10}) + 2 π 1) = 2 - sin (π - arcsin (\frac{2 + 3 6}{10}) + 2 π 1)

整理后得

- 1.06515 \dots = 1.06515 \dots

\Rightarrow 假

x = arcsin (- \frac{- 2 + 3 6}{10}) + 2 πn, x = arcsin (\frac{2 + 3 6}{10}) + 2 πn

以小数形式表示解

x = - 0.56432 \dots + 2 πn, x = 1.20782 \dots + 2 πn

作图

流行的例子

cos(2x)=2-3sin(x)

cos (2 x) = 2 - 3 sin (x)

arcsin(x)+arcsin(1-x)=arccos(x)

arcsin (x) + arcsin (1 - x) = arccos (x)

3cos^2(x)+1=4sin(x)

3 cos^{2} (x) + 1 = 4 sin (x)

[2sin(4x)-1]*[1+tan(x)]=0

[2 sin (4 x) - 1] \cdot [1 + tan (x)] = 0

cos^2(x)=2cos(x)

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