Question

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

• A. $\frac{7\pi }{8},\frac{11\pi }{8}$
• B. $\frac{9\pi }{8},\frac{13\pi }{8}$
• C. $\frac{11\pi }{8},\frac{15\pi }{8}$
• D. $\frac{15\pi }{8},\frac{19\pi }{8}$

Answer 1

Answer: (C)

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

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

$⟹2\pi <2x<4\pi$ ...... $\left(i\right)$

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

$⟹\sin 2x=-\cos 2x$

$⟹\frac{\sin 2x}{\cos 2x}=-1$

$⟹\tan 2x=-1$

$⟹2x={\tan }^{-1}(-1)$

Now, we know that

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

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

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

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

$⟹x=\frac{11\pi }{8},\frac{15\pi }{8}$

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