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If $0\leq x\leq \large\frac{\pi}{2}$ and $81^{\large \sin^2x}+81^{\large\cos^2x}=30$ then $x$ is equal to

$(a)\;\large\frac{\pi}{6}$$\qquad(b)\;\large\frac{\pi}{2}\qquad(c)\;\large\frac{\pi}{4}\qquad$$(d)\;0$

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1 Answer

Step 1:
$81^{\large\sin^2x}+81^{\large\cos^2x}=30$
$81^{\large\sin^2x}+81^{1-\large\sin^2x}=30$
$81\sin^2x\large\frac{81}{81^{\Large\sin^2x}}$$=30$
Let $81^{\large\sin^2x}=y$
$y+\large\frac{81}{y}$$=30$
$y^2-30y+81=0$
$(y-27)(y-3)=0$
$y=27$ (or) $y=3$
Step 2:
$y=27$
$81^{\large\sin^2x}=27$
$3^{\large 4\sin^2x}=3^3$
$\sin^2x=\large\frac{3}{4}$
$\sin x=\large\frac{\sqrt 3}{2}$
$x=\large\frac{\pi}{3}$
Step 3:
$y=3$
$81^{\large\sin^2x}=3$
$3^{\large 4\sin^2x}=3$
$4\sin^2x=1$
$\sin x=\large\frac{1}{2}$
$x=\large\frac{\pi}{6}$
Hence (a) is the correct answer.
answered Nov 19, 2013 by sreemathi.v
 

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