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受欢迎的三角函数 >

solvefor x,Rf=e^{cos(2x)}

解答

求解

x, R f = e^{c o s (2 x)}

解答

x = \frac{arccos ( ln ( R f ) )}{2} + πn, x = - \frac{arccos ( ln ( R f ) )}{2} + πn

求解步骤

R f = e^{c o s (2 x)}

交换两边

e^{c o s (2 x)} = R f

使用指数运算法则

e^{c o s (2 x)} = R f

若

f (x) = g (x)

，则

ln (f (x)) = ln (g (x))

ln (e^{c o s (2 x)}) = ln (R f)

使用对数计算法则:

ln (e^{a}) = a

ln (e^{c o s (2 x)}) = cos (2 x)

cos (2 x) = ln (R f)

cos (2 x) = ln (R f)

使用反三角函数性质

cos (2 x) = ln (R f)

cos (2 x) = ln (R f)

的通解

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

2 x = arccos (ln (R f)) + 2 πn, 2 x = - arccos (ln (R f)) + 2 πn

2 x = arccos (ln (R f)) + 2 πn, 2 x = - arccos (ln (R f)) + 2 πn

解

2 x = arccos (ln (R f)) + 2 πn : x = \frac{arccos ( ln ( R f ) )}{2} + πn

2 x = arccos (ln (R f)) + 2 πn

两边除以

2

2 x = arccos (ln (R f)) + 2 πn

两边除以

2

\frac{2 x}{2} = \frac{arccos ( ln ( R f ) )}{2} + \frac{2 πn}{2}

化简

x = \frac{arccos ( ln ( R f ) )}{2} + πn

x = \frac{arccos ( ln ( R f ) )}{2} + πn

解

2 x = - arccos (ln (R f)) + 2 πn : x = - \frac{arccos ( ln ( R f ) )}{2} + πn

2 x = - arccos (ln (R f)) + 2 πn

两边除以

2

2 x = - arccos (ln (R f)) + 2 πn

两边除以

2

\frac{2 x}{2} = - \frac{arccos ( ln ( R f ) )}{2} + \frac{2 πn}{2}

化简

x = - \frac{arccos ( ln ( R f ) )}{2} + πn

x = - \frac{arccos ( ln ( R f ) )}{2} + πn

x = \frac{arccos ( ln ( R f ) )}{2} + πn, x = - \frac{arccos ( ln ( R f ) )}{2} + πn

作图

流行的例子

sin(2θ)= 12/13

sin (2 θ) = \frac{12}{13}

sin^2(θ)+sin(θ)=2

sin^{2} (θ) + sin (θ) = 2

7sin^2(θ)-3sin(θ)=4

7 sin^{2} (θ) - 3 sin (θ) = 4

- 5 - 4 tan (θ) = - 1

sin (θ) = - \frac{7}{10}