Question

The trigonometric equation is-
$\sin x+3\sin 2x+\sin 3x=\cos x+3\cos 2x+\cos 3x$
when $x$ lies in first four quadrants. It means $xϵ[0,2\pi ]$, then-

How many solutions are there-

• A. $2$
• B. $3$
• C. $4$
• D. $5$

Answer 1

The correct option is C $4$

$\sin x+3\sin 2x+\sin 3x=\cos x+3\cos 2x+\cos 3x$

$2\sin 2x\cos x+3\sin 2x=2\cos 2x\cos x+3\cos 2x$

$\Rightarrow \left(\sin 2x-\cos 2x)\right(2\cos x+3)=0$

$\Rightarrow \tan 2x=1,\cos x=-\frac{3}{2}$ (not possible)

Since, $0\le x\le 2\pi$

$0\le 2x\le 4\pi$

Hence, there are a number of solutions $=4$