Question

Solve :${\log }_{\frac{-x^2-6x}{10}}(\sin 3x+\sin x)={\log }_{\frac{-x^2-6x}{10}}\left(\sin 2x\right)$

• A. $x=\frac{5\pi }{3}$
• B. $x=-\frac{5\pi }{3}$
• C. $x=-\frac{2\pi }{3}$
• D. $x=-\frac{4\pi }{3}$

Answer 1

The correct option is B $x=-\frac{5\pi }{3}$

(i) $-(x^2+6x)>0\Rightarrow x(x+6)<0\Rightarrow x\in (-6,0)$
Also $-(x^2+6x)\ne 10\Rightarrow x^2+6x+10\ne 0,$ This true for all $R$
(ii) $\sin 3x+\sin x,\sin 2x>0$
Now, $\sin 3x+\sin x=\sin 2x$
$\Rightarrow 2\sin 2x.\cos x=\sin x\Rightarrow 4\sin x.{\cos }^{2}x=\sin x$
$\Rightarrow {\cos }^{2}x=\frac{1}{4}\Rightarrow \cos x=\pm \frac{1}{2}$
Following condition (i) and (ii) the solution is, $x=-\frac{5\pi }{3}$