What is the general solution for solvefor t,y=cos(2t) ?

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

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solvefor t,y=cos(2t)

solve for

t, y = cos (2 t)

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

Solution steps

y = cos (2 t)

Switch sides

cos (2 t) = y

Apply trig inverse properties

cos (2 t) = y

General solutions for

cos (2 t) = y

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

2 t = arccos (y) + 2 πn, 2 t = - arccos (y) + 2 πn

2 t = arccos (y) + 2 πn, 2 t = - arccos (y) + 2 πn

Solve

2 t = arccos (y) + 2 πn : t = \frac{arccos ( y )}{2} + πn

2 t = arccos (y) + 2 πn

Divide both sides by

2

2 t = arccos (y) + 2 πn

Divide both sides by

2

\frac{2 t}{2} = \frac{arccos ( y )}{2} + \frac{2 πn}{2}

Simplify

t = \frac{arccos ( y )}{2} + πn

t = \frac{arccos ( y )}{2} + πn

Solve

2 t = - arccos (y) + 2 πn : t = - \frac{arccos ( y )}{2} + πn

2 t = - arccos (y) + 2 πn

Divide both sides by

2

2 t = - arccos (y) + 2 πn

Divide both sides by

2

\frac{2 t}{2} = - \frac{arccos ( y )}{2} + \frac{2 πn}{2}

Simplify

t = - \frac{arccos ( y )}{2} + πn

t = - \frac{arccos ( y )}{2} + πn

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

cos (3 x) = \frac{1}{5}

cos^{2} (x) - 5 cos (x) + 6 = 0

cot^{2} (\frac{x}{2}) = 3, 0, 0 \leq x \leq 360

2 sin^{2} (x) + 5 sin (x) - 3 = 0, 0 \leq x \leq 2 π

sin (x) = \frac{123 sin ( 7 0 ^{\circ} )}{184.5}

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