Question

Solve the equation : $4^{sin2x+2co{s}^{2}x}$ $+$ $4^{1-sin2x+2si{n}^{2}x}$ $=$ 65

Answer 1

$4^{\sin 2x+2{\cos }^{2}x}+4^{1-\sin 2x+2{\sin }^{2}x}=65$

${\sin }^{2}x=1-{\cos }^{2}x$

$\Rightarrow 2{\sin }^{2}x=2-2{\cos }^{2}x$

$\Rightarrow 4^{\sin 2x+2{\cos }^{2}x}+4^{3-(\sin 2x+2{\cos }^{2}x)}=65$

$\Rightarrow 4^{\sin 2x+2{\cos }^{2}x}+\frac{4^3}{4^{\sin 2x+2{\cos }^{2}x}}=65$

using $4^{\sin 2x+2{\cos }^{2}x}=t$

$\Rightarrow t+\frac{64}{t}=65$

$t^2=64=65t$

$\Rightarrow t^2-65t+64=0$

$\Rightarrow (t-1)(t-64)=0$

$\Rightarrow t=1,64$

at $t=1$ at $t=64$

$\Rightarrow 4^{\sin 2x+2{\cos }^{2}x}=1$ $4^{\sin 2x+2{\cos }^{2}x}=4^3$

$\Rightarrow \sin 2x+2{\cos }^{2}x=0$ $\Rightarrow \sin 2x+1+\cos 2x=3$

$\Rightarrow \sin 2x+1+\cos 2x=0$ $\Rightarrow \sin 2x+\cos 2x=2$

$\Rightarrow \cos (2x+\frac{\pi }{4})=\cos \frac{3\pi }{4}$ but maximum value of $\sin 2x+\cos 2x$ is $\sqrt{2}$

$\Rightarrow 2x+\frac{\pi }{4}=2x\pi \pm \frac{3\pi }{4}$ $\Rightarrow$ No solution for x

$\Rightarrow x=n\pi \pm \frac{3\pi }{8}-\frac{\pi }{6}$ $\Rightarrow$ Only solution for x is $x=n\pi \pm \frac{3\pi }{8}-\frac{\pi }{8}$.