A) \[\frac{\pi }{6}\]
B) \[\frac{\pi }{2}\]
C) \[\frac{\pi }{4}\]
D) \[\frac{3\pi }{4}\]
Correct Answer: A
Solution :
\[{{81}^{{{\sin }^{2}}x}}+{{81}^{{{\cos }^{2}}x}}=30\] \[\Rightarrow \] \[{{81}^{{{\sin }^{2}}x}}+{{81}^{1-{{\sin }^{2}}x}}=30\] \[\Rightarrow \] \[{{81}^{{{\sin }^{2}}x}}+\frac{81}{{{81}^{1-{{\sin }^{2}}x}}}=30\] 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, \[y-3=0\] \[y=3\] \[{{81}^{{{\sin }^{2}}x}}=3\] \[{{3}^{4{{\sin }^{2}}x}}={{3}^{1}}\] \[\Rightarrow \] \[4\,{{\sin }^{2}}x=1\] \[\Rightarrow \] \[\sin x=\frac{1}{2}=\sin \frac{\pi }{6}\] \[\Rightarrow \] \[x=\frac{\pi }{6}\]
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