Question

Solve the following equations.
$2^{si{n}^{2}x}+4.2^{co{s}^{2}x}=6$

Answer 1

$\Rightarrow 2^{{\sin }^{2}x}+{4.2}^{1-{\sin }^{2}x}=6$

$\Rightarrow 2^{{\sin }^{2}x}+\frac{4.2}{2^{{\sin }^{2}x}}=6$

$\Rightarrow 2^{{\sin }^{2}x}+\frac{8}{2^{{\sin }^{2}x}}=6$

Let $2^{{\sin }^{2}x}=y$, then

$y+\frac{8}{y}=6$

$\Rightarrow y^2+8=6y$

$\Rightarrow y^2-6y+8=0$

$\Rightarrow y^2-4y-2y+8=0$

$\Rightarrow (y-4)(y-2)=0$

either $y-4=0\Rightarrow y=4\Rightarrow 2^{{\sin }^{2}x}=4\Rightarrow 2^{{\sin }^{2}x}=2^2$

$\Rightarrow {\sin }^{2}x=2$......(1)

or

$y-2=0\Rightarrow y=2\Rightarrow 2^{{\sin }^{2}x}=2\Rightarrow 2^{{\sin }^{2}x}=2$

$\Rightarrow {\sin }^{2}x=1$.......(ii)

(i) cannot here a sol in $x$ as $-1\le \sin x\le 1$ and

(ii) $\sin x=1\Rightarrow x=(4n+1).\frac{\pi }{2}$ for every integer $n$

or $\sin x=-1\Rightarrow x=(4n-1)\frac{\pi }{2}$ for every integer $n$