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

Given : $\pi< x< 2\pi$

$\implies 2\pi<2x<4\pi$ ...... $(i)$

Also, $\sin 2x+\cos 2x=0$ ..... $(ii)$ [Given]

$\implies \sin 2x=-\cos 2x$

$\implies \dfrac{\sin 2x}{\cos 2x}=-1$

$\implies \tan 2x=-1$

$\implies 2x=\tan^{-1}(-1)$

Now, we know that

$\tan \dfrac{3\pi}{4}=\tan \dfrac{7\pi}{4}=\tan \dfrac{11\pi}{4}=...... =\tan \dfrac{(4n+3)\pi}{4}=-1$

So, the possible values for $2x$ are $\dfrac{3\pi}{4}, \dfrac{7\pi}{4}, \dfrac{11\pi}{4},.....,\dfrac{(4n+3)\pi}{4}$

But, from $(i)$ we get the values for $2x$ as $\dfrac{11\pi}{4}$ and $\dfrac{15 \pi}{4}$.

$\therefore 2x=\dfrac{11\pi}{4}, \dfrac{15\pi}{4}$

$\implies x=\dfrac{11\pi}{8}, \dfrac{15\pi}{8}$

$\therefore$ Principal solutions for $(ii)$ are $\dfrac{11\pi}{8}$ and $\dfrac{15\pi}{8}$.