Question

Number of solutions of the trignometric equation ${\cos }^2(\frac{\pi }{4}(cosX+sinX))$ - ${\tan }^2(X+\frac{\pi }{4}{\tan }^{2}X)$ = 1 in [ - 2$\pi$, 2$\pi$] is

Answer 1

$co{s}^2\left(\frac{\pi }{4}\right(cosx+sinx\left)\right)-ta{n}^2(x+\frac{\pi }{4}ta{n}^{2}x)=1$

$\to co{s}^2\left(\frac{\pi }{4}\right(cosx+sinx\left)\right)=1+ta{n}^2(x+\frac{\pi }{4}ta{n}^{2}x)$

$\to co{s}^2\left(\frac{\pi }{4}\right(cosx+sinx\left)\right)=se{c}^2(x+\frac{\pi }{4}ta{n}^{2}x)$

$\to co{s}^2\left(\frac{\pi }{4}\right(cosx+sinx\left)\right)\times co{s}^2(x+\frac{\pi }{4}ta{n}^{2}x)=1$

so,

$\to co{s}^2\left(\frac{\pi }{4}\right(cosx+sinx\left)\right)=1$ and $co{s}^2(x+\frac{\pi }{4}ta{n}^{2}x)=1$

or

$\to co{s}^2\left(\frac{\pi }{4}\right(cosx+sinx\left)\right)=-1$ and $co{s}^2(x+\frac{\pi }{4}ta{n}^{2}x)=-1$

Thus,

$\frac{\pi }{4}(cosx+sinx)=n\pi$ and $x+\frac{\pi }{4}ta{n}^{2}x=n\pi$ where $n=-2,-1,0,1,2$

n=0,

$\to \frac{\pi }{4}(cosx+sinx)=0$ and $x+\frac{\pi }{4}ta{n}^{2}x=0$

on solving,

$x=-\frac{\pi }{4}$

n=1

$\to \frac{\pi }{4}(cosx+sinx)=\pi$

$\to (cosx+sinx)=4$ (which is not possible so as for other cases)

so only one value of x is satisfied