What is the general solution for cos(2a)= 2/3 ,sin(a) ?

The general solution for cos(2a)= 2/3 ,sin(a) is No Solution for a\in\mathbb{R}

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cos(2a)= 2/3 ,sin(a)

cos (2 a) = \frac{2}{3}, sin (a)

No Solution for a \in R

Solution steps

cos (2 a) = \frac{2}{3}, sin (a)

Apply trig inverse properties

cos (2 a) = \frac{2}{3}

General solutions for

cos (2 a) = \frac{2}{3}

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

2 a = arccos (\frac{2}{3}) + 2 πn, 2 a = 2 π - arccos (\frac{2}{3}) + 2 πn

2 a = arccos (\frac{2}{3}) + 2 πn, 2 a = 2 π - arccos (\frac{2}{3}) + 2 πn

Solve

2 a = arccos (\frac{2}{3}) + 2 πn : a = \frac{arccos ( \frac{2}{3} )}{2} + πn

2 a = arccos (\frac{2}{3}) + 2 πn

Divide both sides by

2

2 a = arccos (\frac{2}{3}) + 2 πn

Divide both sides by

2

\frac{2 a}{2} = \frac{arccos ( \frac{2}{3} )}{2} + \frac{2 πn}{2}

Simplify

a = \frac{arccos ( \frac{2}{3} )}{2} + πn

a = \frac{arccos ( \frac{2}{3} )}{2} + πn

Solve

2 a = 2 π - arccos (\frac{2}{3}) + 2 πn : a = π - \frac{arccos ( \frac{2}{3} )}{2} + πn

2 a = 2 π - arccos (\frac{2}{3}) + 2 πn

Divide both sides by

2

2 a = 2 π - arccos (\frac{2}{3}) + 2 πn

Divide both sides by

2

\frac{2 a}{2} = \frac{2 π}{2} - \frac{arccos ( \frac{2}{3} )}{2} + \frac{2 πn}{2}

Simplify

a = π - \frac{arccos ( \frac{2}{3} )}{2} + πn

a = π - \frac{arccos ( \frac{2}{3} )}{2} + πn

a = \frac{arccos ( \frac{2}{3} )}{2} + πn, a = π - \frac{arccos ( \frac{2}{3} )}{2} + πn

Solutions for the range

sin (a)

No Solution for a \in R

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What is the general solution for cos(2a)= 2/3 ,sin(a) ?
The general solution for cos(2a)= 2/3 ,sin(a) is No Solution for a\in\mathbb{R}