Solve the equation : $4^{s i n 2 x + 2 c o s^{2} x}$ $+$ $4^{1 - s i n 2 x + 2 s i n^{2} x}$ $=$ 65

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Solution

$4^{sin 2 x + 2 {cos}^{2} x} + 4^{1 - sin 2 x + 2 {sin}^{2} x} = 65$
${sin}^{2} x = 1 - {cos}^{2} x$
$\Rightarrow 2 {sin}^{2} x = 2 - 2 {cos}^{2} x$
$\Rightarrow 4^{sin 2 x + 2 {cos}^{2} x} + 4^{3 - (sin 2 x + 2 {cos}^{2} x)} = 65$
$\Rightarrow 4^{sin 2 x + 2 {cos}^{2} x} + \frac{4^{3}}{4^{sin 2 x + 2 {cos}^{2} x}} = 65$
using $4^{sin 2 x + 2 {cos}^{2} x} = t$
$\Rightarrow t + \frac{64}{t} = 65$
$t^{2} = 64 = 65 t$
$\Rightarrow t^{2} - 65 t + 64 = 0$
$\Rightarrow (t - 1) (t - 64) = 0$
$\Rightarrow t = 1, 64$
at $t = 1$ at $t = 64$
$\Rightarrow 4^{sin 2 x + 2 {cos}^{2} x} = 1$ $4^{sin 2 x + 2 {cos}^{2} x} = 4^{3}$
$\Rightarrow sin 2 x + 2 {cos}^{2} x = 0$ $\Rightarrow sin 2 x + 1 + cos 2 x = 3$
$\Rightarrow sin 2 x + 1 + cos 2 x = 0$ $\Rightarrow sin 2 x + cos 2 x = 2$
$\Rightarrow cos (2 x + \frac{π}{4}) = cos \frac{3 π}{4}$ but maximum value of $sin 2 x + cos 2 x$ is $\sqrt{2}$
$\Rightarrow 2 x + \frac{π}{4} = 2 x π \pm \frac{3 π}{4}$ $\Rightarrow$ No solution for x
$\Rightarrow x = n π \pm \frac{3 π}{8} - \frac{π}{6}$ $\Rightarrow$ Only solution for x is $x = n π \pm \frac{3 π}{8} - \frac{π}{8}$ .

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