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受欢迎的三角函数 >

sin(2θ)=(-3)/4 ,(3pi)/2 <= 2θ<= 2pi

解答

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

解答

θ \in R 无解

求解步骤

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

使用分式法则:

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

sin (2 θ) = - \frac{3}{4}

使用反三角函数性质

sin (2 θ) = - \frac{3}{4}

sin (2 θ) = - \frac{3}{4}

的通解

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

2 θ = arcsin (- \frac{3}{4}) + 2 πn, 2 θ = π + arcsin (\frac{3}{4}) + 2 πn

2 θ = arcsin (- \frac{3}{4}) + 2 πn, 2 θ = π + arcsin (\frac{3}{4}) + 2 πn

解

2 θ = arcsin (- \frac{3}{4}) + 2 πn : θ = - \frac{arcsin ( \frac{3}{4} )}{2} + πn

2 θ = arcsin (- \frac{3}{4}) + 2 πn

化简

arcsin (- \frac{3}{4}) + 2 πn : - arcsin (\frac{3}{4}) + 2 πn

arcsin (- \frac{3}{4}) + 2 πn

利用以下特性:

arcsin (- x) = - arcsin (x)

arcsin (- \frac{3}{4}) = - arcsin (\frac{3}{4})

= - arcsin (\frac{3}{4}) + 2 πn

2 θ = - arcsin (\frac{3}{4}) + 2 πn

两边除以

2

2 θ = - arcsin (\frac{3}{4}) + 2 πn

两边除以

2

\frac{2 θ}{2} = - \frac{arcsin ( \frac{3}{4} )}{2} + \frac{2 πn}{2}

化简

θ = - \frac{arcsin ( \frac{3}{4} )}{2} + πn

θ = - \frac{arcsin ( \frac{3}{4} )}{2} + πn

解

2 θ = π + arcsin (\frac{3}{4}) + 2 πn : θ = \frac{π}{2} + \frac{arcsin ( \frac{3}{4} )}{2} + πn

2 θ = π + arcsin (\frac{3}{4}) + 2 πn

两边除以

2

2 θ = π + arcsin (\frac{3}{4}) + 2 πn

两边除以

2

\frac{2 θ}{2} = \frac{π}{2} + \frac{arcsin ( \frac{3}{4} )}{2} + \frac{2 πn}{2}

化简

θ = \frac{π}{2} + \frac{arcsin ( \frac{3}{4} )}{2} + πn

θ = \frac{π}{2} + \frac{arcsin ( \frac{3}{4} )}{2} + πn

θ = - \frac{arcsin ( \frac{3}{4} )}{2} + πn, θ = \frac{π}{2} + \frac{arcsin ( \frac{3}{4} )}{2} + πn

在

\frac{3 π}{2} \leq 2 θ \leq 2 π

范围内的解

θ \in R 无解

作图

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