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受欢迎的三角函数 >

solvefor t,f=cos(t_{0}t)

解答

求解

t, f = cos (t_{0} t)

解答

t = \frac{arccos ( f )}{t _{0}} + \frac{2 πn}{t _{0}}, t = - \frac{arccos ( f )}{t _{0}} + \frac{2 πn}{t _{0}}

求解步骤

f = cos (t_{0} t)

交换两边

cos (t_{0} t) = f

使用反三角函数性质

cos (t_{0} t) = f

cos (t_{0} t) = f

的通解

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

t_{0} t = arccos (f) + 2 πn, t_{0} t = - arccos (f) + 2 πn

t_{0} t = arccos (f) + 2 πn, t_{0} t = - arccos (f) + 2 πn

解

t_{0} t = arccos (f) + 2 πn : t = \frac{arccos ( f )}{t _{0}} + \frac{2 πn}{t _{0}}; t_{0} \neq = 0

t_{0} t = arccos (f) + 2 πn

两边除以

t_{0}; t_{0} \neq = 0

t_{0} t = arccos (f) + 2 πn

两边除以

t_{0}; t_{0} \neq = 0

\frac{t _{0} t}{t _{0}} = \frac{arccos ( f )}{t _{0}} + \frac{2 πn}{t _{0}}; t_{0} \neq = 0

化简

t = \frac{arccos ( f )}{t _{0}} + \frac{2 πn}{t _{0}}; t_{0} \neq = 0

t = \frac{arccos ( f )}{t _{0}} + \frac{2 πn}{t _{0}}; t_{0} \neq = 0

解

t_{0} t = - arccos (f) + 2 πn : t = - \frac{arccos ( f )}{t _{0}} + \frac{2 πn}{t _{0}}; t_{0} \neq = 0

t_{0} t = - arccos (f) + 2 πn

两边除以

t_{0}; t_{0} \neq = 0

t_{0} t = - arccos (f) + 2 πn

两边除以

t_{0}; t_{0} \neq = 0

\frac{t _{0} t}{t _{0}} = - \frac{arccos ( f )}{t _{0}} + \frac{2 πn}{t _{0}}; t_{0} \neq = 0

化简

t = - \frac{arccos ( f )}{t _{0}} + \frac{2 πn}{t _{0}}; t_{0} \neq = 0

t = - \frac{arccos ( f )}{t _{0}} + \frac{2 πn}{t _{0}}; t_{0} \neq = 0

t = \frac{arccos ( f )}{t _{0}} + \frac{2 πn}{t _{0}}, t = - \frac{arccos ( f )}{t _{0}} + \frac{2 πn}{t _{0}}

作图

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