# solve for theta. sin 4theta = square root of 3 sin 2 theta

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### 2 Answers

You need to use double angle identity, such that:

`sin 4 theta = sin 2*(2 theta) => sin 4 theta = 2 sin 2theta*cos 2theta`

Substituting `2 sin 2theta*cos 2theta` in the given equation yields:

`2 sin 2theta*cos 2theta = sqrt3*sin 2theta`

You need to move the terms to one side, such that:

`2 sin 2theta*cos 2theta - sqrt3*sin 2theta = 0`

Factoring out sin 2theta yields:

`sin 2theta*(2cos 2 theta - sqrt 3) = 0 => {(sin 2 theta = 0),(2cos 2 theta - sqrt 3 = 0):}`

`{(2theta = +-cos^(-1)(0) + 2n*pi),(2cos 2 theta = sqrt 3):}`

`{(2theta = +-pi/2 + 2n*pi),(cos 2 theta = (sqrt 3)/2):}`

`{(theta = +-pi/4 + n*pi),(2 theta = +-cos^(-1)(sqrt 3)/2 + 2n*pi):}`

`{(theta = +-pi/4 + n*pi),(2 theta = +-pi/6 + 2n*pi):}`

`{(theta = +-pi/4 + n*pi),(theta = +-pi/12 + n*pi):}`

**Hence, solving for theta the given equation yields **`theta = +-pi/4 + n*pi, theta = +-pi/12 + n*pi.`

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