Question

If ${\log }_3{\left(\frac{2\sin 2x}{3}\right)}^2+1=0,x\in [0,2\pi ],$
then which among the following value(s) of $x$ satisfying the above equation

• A. $x=\frac{\pi }{6}$
• B. $x=\frac{4\pi }{3}$
• C. $x=\frac{\pi }{3}$
• D. $x=\frac{7\pi }{6}$

Answer 1

Answer: (A), (B), (C), (D)

$x=\frac{7\pi }{6}$

${\log }_3{\left(\frac{2\sin 2x}{3}\right)}^2+1=0,\sin 2x\ne 0$
$\Rightarrow {\log }_3{\left(\frac{2\sin 2x}{3}\right)}^2=-1$
$\Rightarrow {\left(\frac{2\sin 2x}{3}\right)}^2=3^{-1}=\frac{1}{3}$
$\Rightarrow \frac{4}{9}{\sin }^{2}2x=\frac{1}{3}\Rightarrow {\sin }^{2}2x=\frac{3}{4}={\left(\frac{\sqrt{3}}{2}\right)}^2\Rightarrow {\sin }^{2}2x={\sin }^2\frac{\pi }{3}\Rightarrow 2x=n\pi \pm \frac{\pi }{3},n\in Z$
$\Rightarrow x=\frac{n\pi }{2}\pm \frac{\pi }{6},n\in Z$