The solutions of the equation sin 2x + cos 2x | vedclass

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Question

(C) Given the equation $\sin 2x + \cos 2x = 0$. Dividing by $\cos 2x$ (assuming $\cos 2x 
eq 0$),we get $	an 2x = -1$. Since $	an 	heta = -1$ impl

Vedclass · Accepted Answer

$\frac{11\pi}{8}, \frac{15\pi}{8}$

Vedclass · Answer

(C) Given the equation $\sin 2x + \cos 2x = 0$. Dividing by $\cos 2x$ (assuming $\cos 2x 
eq 0$),we get $	an 2x = -1$. Since $	an 	heta = -1$ implies $	heta = n\pi - \frac{\pi}{4}$,we have $2x = n\pi - \frac{\pi}{4}$. Thus,$x = \frac{n\pi}{2} - \frac{\pi}{8} = \frac{(4n - 1)\pi}{8}$. For $\pi If $n = 3$,$x = \frac{(12 - 1)\pi}{8} = \frac{11\pi}{8}$. If $n = 4$,$x = \frac{(16 - 1)\pi}{8} = \frac{15\pi}{8}$. Both values lie in the interval $(\pi, 2\pi)$.

The solutions of the equation $\sin 2x + \cos 2x = 0$,where $\pi < x < 2\pi$,are

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