${(81)^{{{\sin }^2}x}} + {(81)^{1 - {{\sin }^2}x}} = 30$ ${(81)^{{{\sin }^2}x}} + {{81} \over {{{(81)}^{{{\sin }^2}x}}}} = 30$ Let ${(81)^{{{\sin }^2}x}} = t$ $t + {{81} \over t} = 30 \Rightarrow {t^2} + 81 = 30t$ ${t^2} - 30t + 81 = 0$ ${t^2} - 27t - 3t + 81 = 0$ $(t - 3)(t - 27) = 0$ $t = 3,27$ ${(81)^{{{\sin }^2}x}} = 3,{3^3}$ ${3^{4{{\sin }^2}x}} = {3^1},{3^3}$ $4{\sin ^2}x = 1,3$ ${\sin ^2}x = {1 \over 4},{3 \over 4}$ in [0, $\pi$ ] sin x > 0 $\sin x = {1 \over 2},{{\sqrt 3 } \over 2}$ $x = {\pi \over 6},{{5\pi } \over 6},{\pi \over 3},{{2\pi } \over 3}$ Number of solution = 4