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受欢迎的三角函数 >

solvefor x,sin(y)=ycos(2x)

解答

求解

x, sin (y) = y cos (2 x)

解答

x = \frac{arccos ( \frac{s i n ( y )}{y} )}{2} + πn, x = - \frac{arccos ( \frac{s i n ( y )}{y} )}{2} + πn

求解步骤

sin (y) = y cos (2 x)

交换两边

y cos (2 x) = sin (y)

两边除以

y; y \neq = 0

y cos (2 x) = sin (y)

两边除以

y; y \neq = 0

\frac{y cos ( 2 x )}{y} = \frac{sin ( y )}{y}; y \neq = 0

化简

cos (2 x) = \frac{sin ( y )}{y}; y \neq = 0

cos (2 x) = \frac{sin ( y )}{y}; y \neq = 0

使用反三角函数性质

cos (2 x) = \frac{sin ( y )}{y}

cos (2 x) = \frac{sin ( y )}{y}

的通解

cos (x) = a \Rightarrow x = arccos (a) + 2 πn, x = - arccos (a) + 2 πn

2 x = arccos (\frac{sin ( y )}{y}) + 2 πn, 2 x = - arccos (\frac{sin ( y )}{y}) + 2 πn

2 x = arccos (\frac{sin ( y )}{y}) + 2 πn, 2 x = - arccos (\frac{sin ( y )}{y}) + 2 πn

解

2 x = arccos (\frac{sin ( y )}{y}) + 2 πn : x = \frac{arccos ( \frac{s i n ( y )}{y} )}{2} + πn

2 x = arccos (\frac{sin ( y )}{y}) + 2 πn

两边除以

2

2 x = arccos (\frac{sin ( y )}{y}) + 2 πn

两边除以

2

\frac{2 x}{2} = \frac{arccos ( \frac{s i n ( y )}{y} )}{2} + \frac{2 πn}{2}

化简

x = \frac{arccos ( \frac{s i n ( y )}{y} )}{2} + πn

x = \frac{arccos ( \frac{s i n ( y )}{y} )}{2} + πn

解

2 x = - arccos (\frac{sin ( y )}{y}) + 2 πn : x = - \frac{arccos ( \frac{s i n ( y )}{y} )}{2} + πn

2 x = - arccos (\frac{sin ( y )}{y}) + 2 πn

两边除以

2

2 x = - arccos (\frac{sin ( y )}{y}) + 2 πn

两边除以

2

\frac{2 x}{2} = - \frac{arccos ( \frac{s i n ( y )}{y} )}{2} + \frac{2 πn}{2}

化简

x = - \frac{arccos ( \frac{s i n ( y )}{y} )}{2} + πn

x = - \frac{arccos ( \frac{s i n ( y )}{y} )}{2} + πn

x = \frac{arccos ( \frac{s i n ( y )}{y} )}{2} + πn, x = - \frac{arccos ( \frac{s i n ( y )}{y} )}{2} + πn

作图

流行的例子

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so l v e f or θ, ρ = sin (φ) sin (θ)

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