If x neq (2n+1) frac{pi}{4},then the general | vedclass

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(C) Given the equation: $\cos x + \cos 3x = \sin x + \sin 3x$. Using the sum-to-product formulas $\cos A + \cos B = 2 \cos \frac{A+B}{2} \cos \frac{A-

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$\frac{n \pi}{2} + \frac{\pi}{8}$

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(C) Given the equation: $\cos x + \cos 3x = \sin x + \sin 3x$. Using the sum-to-product formulas $\cos A + \cos B = 2 \cos \frac{A+B}{2} \cos \frac{A-B}{2}$ and $\sin A + \sin B = 2 \sin \frac{A+B}{2} \cos \frac{A-B}{2}$: $2 \cos \frac{x+3x}{2} \cos \frac{x-3x}{2} = 2 \sin \frac{x+3x}{2} \cos \frac{x-3x}{2}$ $2 \cos 2x \cos(-x) = 2 \sin 2x \cos(-x)$ Since $\cos(-x) = \cos x$,we have $2 \cos 2x \cos x = 2 \sin 2x \cos x$. This implies $2 \cos x (\cos 2x - \sin 2x) = 0$. Case $1$: $\cos x = 0$,which gives $x = (2n+1) \frac{\pi}{2}$. However,the condition $x 
eq (2n+1) \frac{\pi}{4}$ is given. Case $2$: $\cos 2x - \sin 2x = 0$,which means $	an 2x = 1$. $2x = n \pi + \frac{\pi}{4}$ $x = \frac{n \pi}{2} + \frac{\pi}{8}$.

If $x \neq (2n+1) \frac{\pi}{4}$,then the general solution of $\cos x + \cos 3x = \sin x + \sin 3x$ is

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