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受欢迎的三角函数 >

3sin(2x)=5cos^2(2x)

解答

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

解答

x = \frac{0.83908 \dots}{2} + πn, x = \frac{π}{2} - \frac{0.83908 \dots}{2} + πn

+1

度数

x = 24.03795 \dots^{\circ} + 18 0^{\circ} n, x = 65.96204 \dots^{\circ} + 18 0^{\circ} n

求解步骤

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

两边减去

5 cos^{2} (2 x)

3 sin (2 x) - 5 cos^{2} (2 x) = 0

使用三角恒等式改写

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

= 3 sin (2 x) - 5 (1 - sin^{2} (2 x))

- (1 - sin^{2} (2 x)) \cdot 5 + 3 sin (2 x) = 0

用替代法求解

- (1 - sin^{2} (2 x)) \cdot 5 + 3 sin (2 x) = 0

令

: sin (2 x) = u

- (1 - u^{2}) \cdot 5 + 3 u = 0

- (1 - u^{2}) \cdot 5 + 3 u = 0 : u = \frac{- 3 + 109}{10}, u = \frac{- 3 - 109}{10}

- (1 - u^{2}) \cdot 5 + 3 u = 0

展开

- (1 - u^{2}) \cdot 5 + 3 u : - 5 + 5 u^{2} + 3 u

- (1 - u^{2}) \cdot 5 + 3 u

= - 5 (1 - u^{2}) + 3 u

乘开

- 5 (1 - u^{2}) : - 5 + 5 u^{2}

- 5 (1 - u^{2})

使用分配律:

a (b - c) = ab - a c

a = - 5, b = 1, c = u^{2}

= - 5 \cdot 1 - (- 5) u^{2}

使用加减运算法则

- (- a) = a

= - 5 \cdot 1 + 5 u^{2}

数字相乘:

5 \cdot 1 = 5

= - 5 + 5 u^{2}

= - 5 + 5 u^{2} + 3 u

- 5 + 5 u^{2} + 3 u = 0

改写成标准形式

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

5 u^{2} + 3 u - 5 = 0

使用求根公式求解

5 u^{2} + 3 u - 5 = 0

二次方程求根公式:

若

a = 5, b = 3, c = - 5

u_{1, 2} = \frac{- 3 \pm 3 ^{2} - 4 \cdot 5 ( - 5 )}{2 \cdot 5}

u_{1, 2} = \frac{- 3 \pm 3 ^{2} - 4 \cdot 5 ( - 5 )}{2 \cdot 5}

3^{2} - 4 \cdot 5 (- 5) = 109

3^{2} - 4 \cdot 5 (- 5)

使用法则

- (- a) = a

= 3^{2} + 4 \cdot 5 \cdot 5

数字相乘:

4 \cdot 5 \cdot 5 = 100

= 3^{2} + 100

3^{2} = 9

= 9 + 100

数字相加:

9 + 100 = 109

= 109

u_{1, 2} = \frac{- 3 \pm 109}{2 \cdot 5}

将解分隔开

u_{1} = \frac{- 3 + 109}{2 \cdot 5}, u_{2} = \frac{- 3 - 109}{2 \cdot 5}

u = \frac{- 3 + 109}{2 \cdot 5} : \frac{- 3 + 109}{10}

\frac{- 3 + 109}{2 \cdot 5}

数字相乘:

2 \cdot 5 = 10

= \frac{- 3 + 109}{10}

u = \frac{- 3 - 109}{2 \cdot 5} : \frac{- 3 - 109}{10}

\frac{- 3 - 109}{2 \cdot 5}

数字相乘:

2 \cdot 5 = 10

= \frac{- 3 - 109}{10}

二次方程组的解是:

u = \frac{- 3 + 109}{10}, u = \frac{- 3 - 109}{10}

u = sin (2 x)

代回

sin (2 x) = \frac{- 3 + 109}{10}, sin (2 x) = \frac{- 3 - 109}{10}

sin (2 x) = \frac{- 3 + 109}{10}, sin (2 x) = \frac{- 3 - 109}{10}

sin (2 x) = \frac{- 3 + 109}{10} : x = \frac{arcsin ( \frac{- 3 + 109}{10} )}{2} + πn, x = \frac{π}{2} - \frac{arcsin ( \frac{- 3 + 109}{10} )}{2} + πn

sin (2 x) = \frac{- 3 + 109}{10}

使用反三角函数性质

sin (2 x) = \frac{- 3 + 109}{10}

sin (2 x) = \frac{- 3 + 109}{10}

的通解

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

2 x = arcsin (\frac{- 3 + 109}{10}) + 2 πn, 2 x = π - arcsin (\frac{- 3 + 109}{10}) + 2 πn

2 x = arcsin (\frac{- 3 + 109}{10}) + 2 πn, 2 x = π - arcsin (\frac{- 3 + 109}{10}) + 2 πn

解

2 x = arcsin (\frac{- 3 + 109}{10}) + 2 πn : x = \frac{arcsin ( \frac{- 3 + 109}{10} )}{2} + πn

2 x = arcsin (\frac{- 3 + 109}{10}) + 2 πn

两边除以

2

2 x = arcsin (\frac{- 3 + 109}{10}) + 2 πn

两边除以

2

\frac{2 x}{2} = \frac{arcsin ( \frac{- 3 + 109}{10} )}{2} + \frac{2 πn}{2}

化简

x = \frac{arcsin ( \frac{- 3 + 109}{10} )}{2} + πn

x = \frac{arcsin ( \frac{- 3 + 109}{10} )}{2} + πn

解

2 x = π - arcsin (\frac{- 3 + 109}{10}) + 2 πn : x = \frac{π}{2} - \frac{arcsin ( \frac{- 3 + 109}{10} )}{2} + πn

2 x = π - arcsin (\frac{- 3 + 109}{10}) + 2 πn

两边除以

2

2 x = π - arcsin (\frac{- 3 + 109}{10}) + 2 πn

两边除以

2

\frac{2 x}{2} = \frac{π}{2} - \frac{arcsin ( \frac{- 3 + 109}{10} )}{2} + \frac{2 πn}{2}

化简

x = \frac{π}{2} - \frac{arcsin ( \frac{- 3 + 109}{10} )}{2} + πn

x = \frac{π}{2} - \frac{arcsin ( \frac{- 3 + 109}{10} )}{2} + πn

x = \frac{arcsin ( \frac{- 3 + 109}{10} )}{2} + πn, x = \frac{π}{2} - \frac{arcsin ( \frac{- 3 + 109}{10} )}{2} + πn

sin (2 x) = \frac{- 3 - 109}{10} :

无解

sin (2 x) = \frac{- 3 - 109}{10}

- 1 \leq sin (x) \leq 1

无解

合并所有解

x = \frac{arcsin ( \frac{- 3 + 109}{10} )}{2} + πn, x = \frac{π}{2} - \frac{arcsin ( \frac{- 3 + 109}{10} )}{2} + πn

以小数形式表示解

x = \frac{0.83908 \dots}{2} + πn, x = \frac{π}{2} - \frac{0.83908 \dots}{2} + πn

作图

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