What is the general solution for solvefor t,f=cos(bt) ?

The general solution for solvefor t,f=cos(bt) is t=(arccos(f))/b+(2pin)/b ,t=-(arccos(f))/b+(2pin)/b

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solvefor t,f=cos(bt)

solve for

t, f = cos (b t)

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

Solution steps

f = cos (b t)

Switch sides

cos (b t) = f

Apply trig inverse properties

cos (b t) = f

General solutions for

cos (b t) = f

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

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

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

Solve

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

b t = arccos (f) + 2 πn

Divide both sides by

b; b \neq = 0

b t = arccos (f) + 2 πn

Divide both sides by

b; b \neq = 0

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

Simplify

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

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

Solve

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

b t = - arccos (f) + 2 πn

Divide both sides by

b; b \neq = 0

b t = - arccos (f) + 2 πn

Divide both sides by

b; b \neq = 0

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

Simplify

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

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

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

3 cos (x) + sin (x) = 0

3 sec (2 x) = 2

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

sin (6 x) = - \frac{1}{2}

2 sin^{2} (θ) + 1 = 3 sin (θ)

What is the general solution for solvefor t,f=cos(bt) ?
The general solution for solvefor t,f=cos(bt) is t=(arccos(f))/b+(2pin)/b ,t=-(arccos(f))/b+(2pin)/b