What is the general solution for solvefor t,f=cos^2(at) ?

The general solution for solvefor t,f=cos^2(at) is

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solvefor t,f=cos^2(at)

Solution

solve for

t, f = cos^{2} (a t)

Solution

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

Solution steps

f = cos^{2} (a t)

Switch sides

cos^{2} (a t) = f

Solve by substitution

cos^{2} (a t) = f

Let:

cos (a t) = u

u^{2} = f

u^{2} = f : u = f, u = - f

u^{2} = f

For

x^{2} = f (a)

the solutions are

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

u = f, u = - f

Substitute back

u = cos (a t)

cos (a t) = f, cos (a t) = - f

cos (a t) = f, cos (a t) = - f

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

cos (a t) = f

Apply trig inverse properties

cos (a t) = f

General solutions for

cos (a t) = f

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

Solve

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

a t = arccos (f) + 2 πn

Divide both sides by

a; a \neq = 0

a t = arccos (f) + 2 πn

Divide both sides by

a; a \neq = 0

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

Simplify

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

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

Solve

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

a t = - arccos (f) + 2 πn

Divide both sides by

a; a \neq = 0

a t = - arccos (f) + 2 πn

Divide both sides by

a; a \neq = 0

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

Simplify

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}

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

cos (a t) = - f

Apply trig inverse properties

cos (a t) = - f

General solutions for

cos (a t) = - f

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

Solve

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

a t = arccos (- f) + 2 πn

Divide both sides by

a; a \neq = 0

a t = arccos (- f) + 2 πn

Divide both sides by

a; a \neq = 0

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

Simplify

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

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

Solve

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

a t = - arccos (- f) + 2 πn

Divide both sides by

a; a \neq = 0

a t = - arccos (- f) + 2 πn

Divide both sides by

a; a \neq = 0

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

Simplify

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}

Combine all the solutions

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

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Frequently Asked Questions (FAQ)

What is the general solution for solvefor t,f=cos^2(at) ?
The general solution for solvefor t,f=cos^2(at) is