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Solve the following equation: sin3theta...

Solve the following equation: ` sin3theta-sintheta=4cos^2theta-2`

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We have `s in3theta-s intheta=4cos^2theta-2`
`2sinthetacos2theta=2cos2theta`
∴`2cos2theta(sintheta−1)=0 `
∴`sintheta=1,cos2theta=0`
∴`theta=nπ+(−1)nπ/2,2theta=nπ+π/2`
or`theta=(2n+1)π/4.`
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