The general solution of cos x + sin x = cos 2 | vedclass

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(A) Given equation: $\cos x + \sin x = \cos 2x + \sin 2x$$\cos x - \cos 2x = \sin 2x - \sin x$Using $\cos C - \cos D = -2 \sin \frac{C+D}{2} \sin \fra

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$1 : 1$

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(A) Given equation: $\cos x + \sin x = \cos 2x + \sin 2x$$\cos x - \cos 2x = \sin 2x - \sin x$Using $\cos C - \cos D = -2 \sin \frac{C+D}{2} \sin \frac{C-D}{2}$ and $\sin C - \sin D = 2 \cos \frac{C+D}{2} \sin \frac{C-D}{2}$:$-2 \sin \frac{3x}{2} \sin \frac{-x}{2} = 2 \cos \frac{3x}{2} \sin \frac{x}{2}$$2 \sin \frac{3x}{2} \sin \frac{x}{2} = 2 \cos \frac{3x}{2} \sin \frac{x}{2}$$2 \sin \frac{x}{2} (\sin \frac{3x}{2} - \cos \frac{3x}{2}) = 0$Case $1$: $\sin \frac{x}{2} = 0 \implies \frac{x}{2} = n\pi \implies x = 2n\pi$. Here $p=2$.Case $2$: $\sin \frac{3x}{2} = \cos \frac{3x}{2} \implies 	an \frac{3x}{2} = 1 = 	an \frac{\pi}{4}$$\frac{3x}{2} = n\pi + \frac{\pi}{4} \implies x = \frac{2n\pi}{3} + \frac{\pi}{6}$. Here $q=2$.Thus, $p:q = 2:2 = 1:1$.

The general solution of $\cos x + \sin x = \cos 2x + \sin 2x$ is $x = np\pi$ or $x = \frac{nq\pi}{3} + \frac{\pi}{6}$ for $n \in \mathbb{Z}$. Then $p : q =$?

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