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Solve: \( \cos ^{2}\left(x+\frac{\pi}{3}\right)=\sin ^{2}\left(\frac{-\pi}{3}-x\right) \)

Answer»

\(cos^2(x + \frac\pi3) = sin^2(\frac{-\pi}3-x)\)

⇒ \(cos^2(x + \frac\pi3) = sin^2(x + \frac\pi3)\)   \((\because sin(-\theta)= -sin\theta)\)

⇒ \(cos^2(x + \frac\pi3)- sin^2(x + \frac \pi3) = 0\)

⇒ \(cos(2(x + \frac \pi3)) =0 = cos\frac\pi2\)   \((\because cos^2\theta -sin^2\theta = cos2\theta)\)

⇒ \(2(x + \frac\pi3) = 2n\pi \pm \frac \pi 2\)

⇒ \(x + \frac \pi3 = n\pi \pm \frac \pi 4\)

⇒ \(x= n\pi \pm \frac\pi4 - \frac\pi3\)

⇒ \(x = n\pi + \frac \pi 4 - \frac\pi3 \; or\; n\pi - \frac\pi4 - \frac\pi3\)

⇒ \(x = n\pi - \frac \pi{12} \;or\; n\pi - \frac{7\pi}{12}, n\in Z\).


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