Question

Solve $2lo{g}_{\left(sinx\right)}2lo{g}_{\left(si{n}^{2}x\right)}a=-1$ stating any condition on ${}^{\prime }{a}^{\prime }$ that may be required for the existence of the solution.

• A. $x=n\pi +(-1{)}^{n}si{n}^{-1}\left\{2^{-\sqrt{-lo{g}_{2}a/2}}\right\}andaϵ(0,1),nϵz$
• B. $x=n\pi \pm {\sin }^{-1}\left\{2^{-\sqrt{-lo{g}_{2}a/2}}\right\}andaϵ(0,1),nϵz$
• C. $x=n\pi +(-1{)}^{n}co{s}^{-1}\left\{2^{-\sqrt{-lo{g}_{2}a/2}}\right\}andaϵ(0,1),nϵz$
• D. $x=n\pi -(-1{)}^{n}si{n}^{-1}\left\{2^{-\sqrt{-lo{g}_{2}a/2}}\right\}andaϵ(0,1),nϵz$

Answer 1

The correct option is A $x=n\pi +(-1{)}^{n}si{n}^{-1}\left\{2^{-\sqrt{-lo{g}_{2}a/2}}\right\}andaϵ(0,1),nϵz$

Given, $\left({\log }_{\sin x}2\right)\left({\log }_{{\sin }^{2}x}a\right)=-1$
$(a>0,\sin x>0,\sin x\ne 1)$
$\Rightarrow \left({\log }_{\sin x}2\right)\left({\log }_{\sin x}a\right)=-2\Rightarrow \left({\log }_{\sin x}2\right)\left(\frac{{\log }_{2}a}{{\log }_2\sin x}\right)=-2$
$\Rightarrow {\log }_{2}a=-2{\left({\log }_2\sin x\right)}^2\Rightarrow {\log }_2\sin x=\pm \sqrt{\frac{-{\log }_{2}a}{2}}$
$\therefore 0<a<1$
${\log }_2\sin x=-\sqrt{\frac{-{\log }_{2}a}{2}}\Rightarrow \sin x=2^{-\sqrt{\frac{-{\log }_{2}a}{2}}}$
$\Rightarrow x=n\pi +{(-1)}^n{\sin }^{-1}\left(2^{-\sqrt{\frac{-{\log }_{2}a}{2}}}\right)$
Where $0<a<1$