• # question_answer If $0\le x\le \pi$ and ${{81}^{{{\sin }^{2}}x}}+{{81}^{{{\cos }^{2}}x}}=30$, then $x$ is equal to : A)  $\frac{\pi }{6}$                                              B)  $\frac{\pi }{2}$ C)                  $\frac{\pi }{4}$                                              D)                  $\frac{3\pi }{4}$

${{81}^{{{\sin }^{2}}x}}+{{81}^{{{\cos }^{2}}x}}=30$ $\Rightarrow$               ${{81}^{{{\sin }^{2}}x}}+{{81}^{1-{{\sin }^{2}}x}}=30$ $\Rightarrow$               ${{81}^{{{\sin }^{2}}x}}+\frac{81}{{{81}^{1-{{\sin }^{2}}x}}}=30$ Let ${{81}^{{{\sin }^{2}}x}}=y$ $\therefore$                  $y+\frac{81}{y}=30$ $\Rightarrow$               ${{y}^{2}}-30y+81=0$ $\Rightarrow$               $(y-27)\,(y-3)=0$ if,            $y-27=0$                                 $y=27$                 ${{81}^{{{\sin }^{2}}x}}=27$                                 ${{3}^{4{{\sin }^{2}}x}}={{3}^{3}}$                                 ${{\sin }^{2}}x=\frac{3}{4}$                                 $\sin x=\frac{\sqrt{3}}{2}=\sin \frac{\pi }{3}$                 $\Rightarrow$               $x=\pi /3$         Now if, $y-3=0$                 $y=3$                 ${{81}^{{{\sin }^{2}}x}}=3$                 ${{3}^{4{{\sin }^{2}}x}}={{3}^{1}}$ $\Rightarrow$               $4\,{{\sin }^{2}}x=1$ $\Rightarrow$                               $\sin x=\frac{1}{2}=\sin \frac{\pi }{6}$ $\Rightarrow$                               $x=\frac{\pi }{6}$