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受欢迎的三角函数 >

solvefor x,f=2cos(3x^2-1)entoncesf

解答

求解

x, f = 2 cos (3 x^{2} - 1) e n t o n ces f

解答

x = \frac{arccos ( \frac{1}{2 e ^{2} c n ^{2} t os} ) + 2 πk + 1}{3}, x = - \frac{arccos ( \frac{1}{2 e ^{2} c n ^{2} t os} ) + 2 πk + 1}{3}, x = \frac{- arccos ( \frac{1}{2 e ^{2} c n ^{2} t os} ) + 2 πk + 1}{3}, x = - \frac{- arccos ( \frac{1}{2 e ^{2} c n ^{2} t os} ) + 2 πk + 1}{3}

求解步骤

f = 2 cos (3 x^{2} - 1) e n t o n ces f

交换两边

2 cos (3 x^{2} - 1) e n t o n ces f = f

两边除以

2 e n t o n ces f; c \neq = 0

2 cos (3 x^{2} - 1) e n t o n ces f = f

两边除以

2 e n t o n ces f; c \neq = 0

\frac{2 cos ( 3 x ^{2} - 1 ) e n t o n ces f}{2 e n t o n ces f} = \frac{f}{2 e n t o n ces f}; c \neq = 0

化简

\frac{2 cos ( 3 x ^{2} - 1 ) e n t o n ces f}{2 e n t o n ces f} = \frac{f}{2 e n t o n ces f}

化简

\frac{2 cos ( 3 x ^{2} - 1 ) e n t o n ces f}{2 e n t o n ces f} : cos (3 x^{2} - 1)

\frac{2 cos ( 3 x ^{2} - 1 ) e n t o n ces f}{2 e n t o n ces f}

2 cos (3 x^{2} - 1) e n t o n ces f = 2 e^{2} c f t os n^{2} cos (3 x^{2} - 1)

2 cos (3 x^{2} - 1) e n t o n ces f

使用指数法则:

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

ee = e^{1 + 1}

= 2 cos (3 x^{2} - 1) n t o n c e^{1 + 1} s f

数字相加:

1 + 1 = 2

= 2 cos (3 x^{2} - 1) n t o n c e^{2} s f

使用指数法则:

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

nn = n^{1 + 1}

= 2 cos (3 x^{2} - 1) t o n^{1 + 1} c e^{2} s f

数字相加:

1 + 1 = 2

= 2 cos (3 x^{2} - 1) t o n^{2} c e^{2} s f

= \frac{2 e ^{2} c f t os n ^{2} cos ( 3 x ^{2} - 1 )}{2 eec f t os nn}

2 e n t o n ces f = 2 e^{2} c f t os n^{2}

2 e n t o n ces f

使用指数法则:

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

ee = e^{1 + 1}

= 2 n t o n c e^{1 + 1} s f

数字相加:

1 + 1 = 2

= 2 n t o n c e^{2} s f

使用指数法则:

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

nn = n^{1 + 1}

= 2 t o n^{1 + 1} c e^{2} s f

数字相加:

1 + 1 = 2

= 2 t o n^{2} c e^{2} s f

= \frac{2 e ^{2} c f t os n ^{2} cos ( 3 x ^{2} - 1 )}{2 e ^{2} c f t os n ^{2}}

数字相除:

\frac{2}{2} = 1

= \frac{e ^{2} c f t os n ^{2} cos ( 3 x ^{2} - 1 )}{e ^{2} c f t os n ^{2}}

约分:

t

= \frac{e ^{2} c f os n ^{2} cos ( 3 x ^{2} - 1 )}{e ^{2} c f os n ^{2}}

约分:

o

= \frac{e ^{2} c f s n ^{2} cos ( 3 x ^{2} - 1 )}{e ^{2} c f s n ^{2}}

约分:

n^{2}

= \frac{e ^{2} c f s cos ( 3 x ^{2} - 1 )}{e ^{2} c f s}

约分:

c

= \frac{e ^{2} f s cos ( 3 x ^{2} - 1 )}{e ^{2} f s}

约分:

e^{2}

= \frac{f s cos ( 3 x ^{2} - 1 )}{f s}

约分:

s

= \frac{f cos ( 3 x ^{2} - 1 )}{f}

约分:

f

= cos (3 x^{2} - 1)

化简

\frac{f}{2 e n t o n ces f} : \frac{1}{2 e ^{2} c t os n ^{2}}

\frac{f}{2 e n t o n ces f}

2 e n t o n ces f = 2 e^{2} c f t os n^{2}

2 e n t o n ces f

使用指数法则:

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

ee = e^{1 + 1}

= 2 n t o n c e^{1 + 1} s f

数字相加:

1 + 1 = 2

= 2 n t o n c e^{2} s f

使用指数法则:

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

nn = n^{1 + 1}

= 2 t o n^{1 + 1} c e^{2} s f

数字相加:

1 + 1 = 2

= 2 t o n^{2} c e^{2} s f

= \frac{f}{2 e ^{2} c f t os n ^{2}}

约分:

f

= \frac{1}{2 e ^{2} c t os n ^{2}}

cos (3 x^{2} - 1) = \frac{1}{2 e ^{2} c t os n ^{2}}; c \neq = 0

cos (3 x^{2} - 1) = \frac{1}{2 e ^{2} c t os n ^{2}}; c \neq = 0

cos (3 x^{2} - 1) = \frac{1}{2 e ^{2} c t os n ^{2}}; c \neq = 0

使用反三角函数性质

cos (3 x^{2} - 1) = \frac{1}{2 e ^{2} c t os n ^{2}}

cos (3 x^{2} - 1) = \frac{1}{2 e ^{2} c t os n ^{2}}

的通解

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

3 x^{2} - 1 = arccos (\frac{1}{2 e ^{2} c t os n ^{2}}) + 2 πk, 3 x^{2} - 1 = - arccos (\frac{1}{2 e ^{2} c t os n ^{2}}) + 2 πk

3 x^{2} - 1 = arccos (\frac{1}{2 e ^{2} c t os n ^{2}}) + 2 πk, 3 x^{2} - 1 = - arccos (\frac{1}{2 e ^{2} c t os n ^{2}}) + 2 πk

解

3 x^{2} - 1 = arccos (\frac{1}{2 e ^{2} c t os n ^{2}}) + 2 πk : x = \frac{arccos ( \frac{1}{2 e ^{2} c n ^{2} t os} ) + 2 πk + 1}{3}, x = - \frac{arccos ( \frac{1}{2 e ^{2} c n ^{2} t os} ) + 2 πk + 1}{3}

3 x^{2} - 1 = arccos (\frac{1}{2 e ^{2} c t os n ^{2}}) + 2 πk

将

1

到右边

3 x^{2} - 1 = arccos (\frac{1}{2 e ^{2} c t os n ^{2}}) + 2 πk

两边加上

1

3 x^{2} - 1 + 1 = arccos (\frac{1}{2 e ^{2} c t os n ^{2}}) + 2 πk + 1

化简

3 x^{2} = arccos (\frac{1}{2 e ^{2} c t os n ^{2}}) + 2 πk + 1

3 x^{2} = arccos (\frac{1}{2 e ^{2} c t os n ^{2}}) + 2 πk + 1

两边除以

3

3 x^{2} = arccos (\frac{1}{2 e ^{2} c t os n ^{2}}) + 2 πk + 1

两边除以

3

\frac{3 x ^{2}}{3} = \frac{arccos ( \frac{1}{2 e ^{2} c t os n ^{2}} )}{3} + \frac{2 πk}{3} + \frac{1}{3}

化简

x^{2} = \frac{arccos ( \frac{1}{2 e ^{2} c t os n ^{2}} )}{3} + \frac{2 πk}{3} + \frac{1}{3}

x^{2} = \frac{arccos ( \frac{1}{2 e ^{2} c t os n ^{2}} )}{3} + \frac{2 πk}{3} + \frac{1}{3}

对于

x^{2} = f (a)

解为

x = f (a), - f (a)

x = \frac{arccos ( \frac{1}{2 e ^{2} c t os n ^{2}} )}{3} + \frac{2 πk}{3} + \frac{1}{3}, x = - \frac{arccos ( \frac{1}{2 e ^{2} c t os n ^{2}} )}{3} + \frac{2 πk}{3} + \frac{1}{3}

化简

\frac{arccos ( \frac{1}{2 e ^{2} c t os n ^{2}} )}{3} + \frac{2 πk}{3} + \frac{1}{3} : \frac{arccos ( \frac{1}{2 e ^{2} c n ^{2} t os} ) + 2 πk + 1}{3}

\frac{arccos ( \frac{1}{2 e ^{2} c t os n ^{2}} )}{3} + \frac{2 πk}{3} + \frac{1}{3}

合并分式

\frac{arccos ( \frac{1}{2 e ^{2} c n ^{2} t os} )}{3} + \frac{2 πk}{3} + \frac{1}{3} : \frac{arccos ( \frac{1}{2 e ^{2} c n ^{2} t os} ) + 2 πk + 1}{3}

使用法则

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{arccos ( \frac{1}{2 e ^{2} c n ^{2} t os} ) + 2 πk + 1}{3}

= \frac{arccos ( \frac{1}{2 e ^{2} c t os n ^{2}} ) + 2 πk + 1}{3}

= \frac{arccos ( \frac{1}{2 e ^{2} c n ^{2} t os} ) + 2 πk + 1}{3}

化简

- \frac{arccos ( \frac{1}{2 e ^{2} c t os n ^{2}} )}{3} + \frac{2 πk}{3} + \frac{1}{3} : - \frac{arccos ( \frac{1}{2 e ^{2} c n ^{2} t os} ) + 2 πk + 1}{3}

- \frac{arccos ( \frac{1}{2 e ^{2} c t os n ^{2}} )}{3} + \frac{2 πk}{3} + \frac{1}{3}

合并分式

\frac{arccos ( \frac{1}{2 e ^{2} c n ^{2} t os} )}{3} + \frac{2 πk}{3} + \frac{1}{3} : \frac{arccos ( \frac{1}{2 e ^{2} c n ^{2} t os} ) + 2 πk + 1}{3}

使用法则

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{arccos ( \frac{1}{2 e ^{2} c n ^{2} t os} ) + 2 πk + 1}{3}

= - \frac{2 πk + arccos ( \frac{1}{2 e ^{2} c n ^{2} t os} ) + 1}{3}

= - \frac{arccos ( \frac{1}{2 e ^{2} c n ^{2} t os} ) + 2 πk + 1}{3}

x = \frac{arccos ( \frac{1}{2 e ^{2} c n ^{2} t os} ) + 2 πk + 1}{3}, x = - \frac{arccos ( \frac{1}{2 e ^{2} c n ^{2} t os} ) + 2 πk + 1}{3}

解

3 x^{2} - 1 = - arccos (\frac{1}{2 e ^{2} c t os n ^{2}}) + 2 πk : x = \frac{- arccos ( \frac{1}{2 e ^{2} c n ^{2} t os} ) + 2 πk + 1}{3}, x = - \frac{- arccos ( \frac{1}{2 e ^{2} c n ^{2} t os} ) + 2 πk + 1}{3}

3 x^{2} - 1 = - arccos (\frac{1}{2 e ^{2} c t os n ^{2}}) + 2 πk

将

1

到右边

3 x^{2} - 1 = - arccos (\frac{1}{2 e ^{2} c t os n ^{2}}) + 2 πk

两边加上

1

3 x^{2} - 1 + 1 = - arccos (\frac{1}{2 e ^{2} c t os n ^{2}}) + 2 πk + 1

化简

3 x^{2} = - arccos (\frac{1}{2 e ^{2} c t os n ^{2}}) + 2 πk + 1

3 x^{2} = - arccos (\frac{1}{2 e ^{2} c t os n ^{2}}) + 2 πk + 1

两边除以

3

3 x^{2} = - arccos (\frac{1}{2 e ^{2} c t os n ^{2}}) + 2 πk + 1

两边除以

3

\frac{3 x ^{2}}{3} = - \frac{arccos ( \frac{1}{2 e ^{2} c t os n ^{2}} )}{3} + \frac{2 πk}{3} + \frac{1}{3}

化简

x^{2} = - \frac{arccos ( \frac{1}{2 e ^{2} c t os n ^{2}} )}{3} + \frac{2 πk}{3} + \frac{1}{3}

x^{2} = - \frac{arccos ( \frac{1}{2 e ^{2} c t os n ^{2}} )}{3} + \frac{2 πk}{3} + \frac{1}{3}

对于

x^{2} = f (a)

解为

x = f (a), - f (a)

x = - \frac{arccos ( \frac{1}{2 e ^{2} c t os n ^{2}} )}{3} + \frac{2 πk}{3} + \frac{1}{3}, x = - - \frac{arccos ( \frac{1}{2 e ^{2} c t os n ^{2}} )}{3} + \frac{2 πk}{3} + \frac{1}{3}

化简

- \frac{arccos ( \frac{1}{2 e ^{2} c t os n ^{2}} )}{3} + \frac{2 πk}{3} + \frac{1}{3} : \frac{- arccos ( \frac{1}{2 e ^{2} c n ^{2} t os} ) + 2 πk + 1}{3}

- \frac{arccos ( \frac{1}{2 e ^{2} c t os n ^{2}} )}{3} + \frac{2 πk}{3} + \frac{1}{3}

合并分式

- \frac{arccos ( \frac{1}{2 e ^{2} c n ^{2} t os} )}{3} + \frac{2 πk}{3} + \frac{1}{3} : \frac{- arccos ( \frac{1}{2 e ^{2} c n ^{2} t os} ) + 2 πk + 1}{3}

使用法则

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{- arccos ( \frac{1}{2 e ^{2} c n ^{2} t os} ) + 2 πk + 1}{3}

= \frac{- arccos ( \frac{1}{2 e ^{2} c t os n ^{2}} ) + 2 πk + 1}{3}

= \frac{- arccos ( \frac{1}{2 e ^{2} c n ^{2} t os} ) + 2 πk + 1}{3}

化简

- - \frac{arccos ( \frac{1}{2 e ^{2} c t os n ^{2}} )}{3} + \frac{2 πk}{3} + \frac{1}{3} : - \frac{- arccos ( \frac{1}{2 e ^{2} c n ^{2} t os} ) + 2 πk + 1}{3}

- - \frac{arccos ( \frac{1}{2 e ^{2} c t os n ^{2}} )}{3} + \frac{2 πk}{3} + \frac{1}{3}

合并分式

- \frac{arccos ( \frac{1}{2 e ^{2} c n ^{2} t os} )}{3} + \frac{2 πk}{3} + \frac{1}{3} : \frac{- arccos ( \frac{1}{2 e ^{2} c n ^{2} t os} ) + 2 πk + 1}{3}

使用法则

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{- arccos ( \frac{1}{2 e ^{2} c n ^{2} t os} ) + 2 πk + 1}{3}

= - \frac{2 πk + 1 - arccos ( \frac{1}{2 e ^{2} c n ^{2} t os} )}{3}

= - \frac{- arccos ( \frac{1}{2 e ^{2} c n ^{2} t os} ) + 2 πk + 1}{3}

x = \frac{- arccos ( \frac{1}{2 e ^{2} c n ^{2} t os} ) + 2 πk + 1}{3}, x = - \frac{- arccos ( \frac{1}{2 e ^{2} c n ^{2} t os} ) + 2 πk + 1}{3}

x = \frac{arccos ( \frac{1}{2 e ^{2} c n ^{2} t os} ) + 2 πk + 1}{3}, x = - \frac{arccos ( \frac{1}{2 e ^{2} c n ^{2} t os} ) + 2 πk + 1}{3}, x = \frac{- arccos ( \frac{1}{2 e ^{2} c n ^{2} t os} ) + 2 πk + 1}{3}, x = - \frac{- arccos ( \frac{1}{2 e ^{2} c n ^{2} t os} ) + 2 πk + 1}{3}

作图

流行的例子

solvefor x,z=tan(x/2)

tan(x/2)=sqrt(3),0<= x<= 2pi