Question

Solve $\sin 2\theta +\cos 2\theta =0$

• A. $\theta =2n\pi -\frac{\pi }{3}$ $n\in Z$,
• B. $\theta =\frac{2n\pi }{3}-\frac{\pi }{4}$, $n\in Z$
• C. $\theta =2n\pi -\frac{\pi }{2}$ $n\in Z$,
• D. $\theta =\frac{2n\pi }{3}-\frac{\pi }{6}$, $n\in Z$

Answer 1

Answer: (C), (D)

The correct options are
C $\theta =2n\pi -\frac{\pi }{2}$ $n\in Z$,
D $\theta =\frac{2n\pi }{3}-\frac{\pi }{6}$, $n\in Z$

We have, $\sin 2\theta +\cos 2\theta =0$

$\Rightarrow \cos 2\theta =-\sin 2\theta$ $=\cos (\frac{\pi }{2}+2\theta )$

$\Rightarrow$ $\theta =2n\pi \pm (\frac{\pi }{2}+2\theta )$ $n\in Z$

Taking a positive sign, we have

$\theta =2n\pi +\frac{\pi }{2}+2\theta$

$=2n\pi -\frac{\pi }{2}$ $nϵZ$

Taking negative sign, we have
$\theta =2n\pi -(\frac{\pi }{2}+2\theta )$ $\Rightarrow$ $\theta =\frac{2n\pi }{3}-\frac{\pi }{6}$, $n\in Z$