Question

$e^{(1+{sin}^{2}x+{sin}^{4}x+....\infty )log2}=16$ then

Answer 1

We have,

$e^{(1+{\sin }^{2}x+{\sin }^{4}x+......\infty )\log 2}=16$

$\Rightarrow (1+{\sin }^{2}x+{\sin }^{4}x+.....\infty )\log 2=\log 16$

Now,

$(1+{\sin }^{2}x+{\sin }^{4}x+.....\infty )$ series is G.P.

So, First term

$a=1$

$r\left(\text{Common}\text{ratio}\right)=\frac{{\sin }^{2}x}{1}$

$r={\sin }^{2}x$

The sum of this series is

$S_{\infty }=\frac{a}{1-r}$

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

$=\frac{1}{{\cos }^{2}x}$

$={\sec }^{2}x$

So,

${\sec }^{2}x\log 2=\log 16$

$\Rightarrow {\sec }^{2}x\log 2=\log 2^4$

$\Rightarrow {\sec }^{2}x\log 2=4\log 2$

Now comparing and we get,

${\sec }^{2}x=4$

${\sec }^{2}x=2^2$

$\sec x=2$

$\mathrm{secx}=sec\frac{\pi }{6}$

$x=\frac{\pi }{6}$

Hence, this is the answer.