A geometry problem by Rajdeep Bharati

Let $y = 81^{\sin^2{x}}$ . Then, we have:

$\begin{aligned} 81^{\sin^2{x}} + 81^{\cos^2{x}} & = 30 \\ 81^{\sin^2{x}} + 81^{1-\sin^2{x}} & = 30 \\ 81^{\sin^2{x}} + \frac{81}{81^{\sin^2{x}}} & = 30 \\ y + \frac{81}{y} & = 30 \\ y^2 - 30 y + 81 & = 0 \\ (y-3)(y-27) & = 0 \end{aligned}$

$\Rightarrow y = \begin{cases} 3^{4\sin^2{x}} = 3^1 & \Rightarrow 4\sin^2{x} = 1 & \Rightarrow \sin{x} = \frac{1}{2} & \Rightarrow x = \frac{\pi}{6} \\ 3^{4\sin^2{x}} = 3^3 & \Rightarrow 4\sin^2{x} = 3 & \Rightarrow \sin{x} = \frac{\sqrt{3}}{2} & \Rightarrow x = \frac{\pi}{3} \end{cases}$

$\Rightarrow x = \boxed{\frac{\pi}{3}, \frac{\pi}{6}}$

We begin with $81^{\sin^2(x)}+81^{\cos^2(x)} = 3^{4\sin^2(x)} + 3^{4\cos^2(x)}=30$ So, essentially, we need to find two powers of $3$ that add to $30$ . A quick check tells us that this must be either $3+3^3$ or $3^3+3$ . Solving these systems: $\begin{cases} 4\sin^2(x) = 3 \\ 4\cos^2(x)=1 \end{cases} \Rightarrow \boxed{x= \dfrac{\pi}{3}}$ $\begin{cases} 4\sin^2(x) = 1 \\ 4\cos^2(x)=3 \end{cases} \Rightarrow \boxed{x= \dfrac{\pi}{6}}$ Hence the answer.

LHS=81^(sin^2 x)+81/(81^(sin^2 x))=30. Let 81^(sin^2 x)=alpha. Then we have alpha+81/alpha=30, i.e., alpha^2-30 alpha +81=0, => (alpha-27)(alpha-3)=0, i.e., 81^(sin^2 x)=27 or 81^(sin^2 x)=3, i.e., sin^2 x=3/4 or sin^2 x=1/2, yielding sin x=+ or - sqrt(3)/2, + or - 1/2, i.e., x=+ or - pi/3, +or - pi/6