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受欢迎的三角函数 >

solvefor t,f=cos(at)

解答

求解

t, f = cos (a t)

解答

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

求解步骤

f = cos (a t)

交换两边

cos (a t) = f

使用反三角函数性质

cos (a t) = f

cos (a t) = f

的通解

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

a t = arccos (f) + 2 πn, a t = - arccos (f) + 2 πn

a t = arccos (f) + 2 πn, a t = - arccos (f) + 2 πn

解

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

a t = arccos (f) + 2 πn

两边除以

a; a \neq = 0

a t = arccos (f) + 2 πn

两边除以

a; a \neq = 0

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

化简

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

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

解

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

a t = - arccos (f) + 2 πn

两边除以

a; a \neq = 0

a t = - arccos (f) + 2 πn

两边除以

a; a \neq = 0

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

化简

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

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

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

作图

流行的例子

sqrt(2)cos(x)+1=0,pi<x<4pi

2 cos (x) + 1 = 0, π < x < 4 π

cos (3 x) = 3

4cos^2(θ)-5cos(θ)=-4cos(θ)+3

4 cos^{2} (θ) - 5 cos (θ) = - 4 cos (θ) + 3

solvefor x,cos(x)=-0.5

so l v e f or x, cos (x) = - 0.5

-2cos(x)=3sin(x)

- 2 cos (x) = 3 sin (x)