The number of solutions of the equation 32^{t | vedclass

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(B) Given equation: $32^{	an^{2} x} + 32^{\sec^{2} x} = 81$Using the identity $\sec^{2} x = 1 + 	an^{2} x$,we get:$32^{	an^{2} x} + 32^{1 + 	an^{2

Vedclass · Accepted Answer

$1$

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(B) Given equation: $32^{	an^{2} x} + 32^{\sec^{2} x} = 81$Using the identity $\sec^{2} x = 1 + 	an^{2} x$,we get:$32^{	an^{2} x} + 32^{1 + 	an^{2} x} = 81$Let $y = 32^{	an^{2} x}$. Then the equation becomes:$y + 32y = 81$$33y = 81$$y = \frac{81}{33} = \frac{27}{11}$So,$32^{	an^{2} x} = \frac{27}{11}$.Taking $\log_{32}$ on both sides:$	an^{2} x = \log_{32} \left(\frac{27}{11}ight)$.Since $0 \leq x \leq \frac{\pi}{4}$,we have $0 \leq 	an^{2} x \leq 1$.Since $1 Thus,there exists exactly one value of $	an x$ in the interval $[0, 1]$,which corresponds to exactly one value of $x$ in $[0, \frac{\pi}{4}]$.Therefore,the number of solutions is $1$.

The number of solutions of the equation $32^{\tan^{2} x} + 32^{\sec^{2} x} = 81$ for $0 \leq x \leq \frac{\pi}{4}$ is:

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