Question

If $-\frac{\pi }{2}<x<\frac{\pi }{2}$ and the sum to infinite number of terms of the series$cosx+\frac{2}{3}cosxsi{n}^{2}x+\frac{4}{9}cosxsi{n}^{4}x+.....$ is finite, then $x$ lies in the set

• A. $-\frac{\pi }{2}<x<\frac{\pi }{2}$
• B. $-\frac{\pi }{6}<x<\frac{\pi }{6}$
• C. $\frac{\pi }{6}<x<\frac{\pi }{2}$
• D. $0<x<\frac{\pi }{2}$

Answer 1

The correct option is A

$-\frac{\pi }{2}<x<\frac{\pi }{2}$

Solution : $cosx+\frac{2}{3}cosxsi{n}^{2}x+\frac{4}{9}cosxsi{n}^{4}x+.....$ is finite

$\frac{cosx}{1-\frac{2}{3}si{n}^{2}x}$

$\frac{3Cosx}{3-2si{n}^{2}x}$

$\frac{3cosx}{1+2(1-si{n}^{2}x)}$

$\frac{3cosx}{1+co{s}^{2}x}$

$-1<\frac{2}{3}si{n}^{2}x<1$

$-3<2si{n}^{2}x<3$

$-\frac{3}{2}<si{n}^{2}x<\frac{3}{2}$

$0<si{n}^{2}x<\frac{3}{2}$

We know that for every value of x we have $0<si{n}^{2}x<1$

Since, $-\frac{\pi }{2}<x<\frac{\pi }{2}$

Then x can take all the values in the interval -

$-\frac{\pi }{2}<x<\frac{\pi }{2}$

$\therefore -\frac{\pi }{2}<x<\frac{\pi }{2}$