What is the general solution for solvefor a,r=cos(2a) ?

The general solution for solvefor a,r=cos(2a) is a=(arccos(r))/2+pin,a=-(arccos(r))/2+pin

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solvefor a,r=cos(2a)

solve for

a, r = cos (2 a)

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

Solution steps

r = cos (2 a)

Switch sides

cos (2 a) = r

Apply trig inverse properties

cos (2 a) = r

General solutions for

cos (2 a) = r

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

2 a = arccos (r) + 2 πn, 2 a = - arccos (r) + 2 πn

2 a = arccos (r) + 2 πn, 2 a = - arccos (r) + 2 πn

Solve

2 a = arccos (r) + 2 πn : a = \frac{arccos ( r )}{2} + πn

2 a = arccos (r) + 2 πn

Divide both sides by

2

2 a = arccos (r) + 2 πn

Divide both sides by

2

\frac{2 a}{2} = \frac{arccos ( r )}{2} + \frac{2 πn}{2}

Simplify

a = \frac{arccos ( r )}{2} + πn

a = \frac{arccos ( r )}{2} + πn

Solve

2 a = - arccos (r) + 2 πn : a = - \frac{arccos ( r )}{2} + πn

2 a = - arccos (r) + 2 πn

Divide both sides by

2

2 a = - arccos (r) + 2 πn

Divide both sides by

2

\frac{2 a}{2} = - \frac{arccos ( r )}{2} + \frac{2 πn}{2}

Simplify

a = - \frac{arccos ( r )}{2} + πn

a = - \frac{arccos ( r )}{2} + πn

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

cos (2 t) + cos (t) = 0

\frac{π}{6} = sin (3 x)

tan (x) = 3 tan (\frac{1}{2} x)

cos (t) = \frac{- 5}{13}, \frac{1}{2} π < t < π

31.91 cos (x) = 20.48

What is the general solution for solvefor a,r=cos(2a) ?
The general solution for solvefor a,r=cos(2a) is a=(arccos(r))/2+pin,a=-(arccos(r))/2+pin