Question

Solve the equation $co{s}^2\left[\frac{\pi }{4}\right(sinx+\sqrt{2}co{s}^{2}x\left)\right]-ta{n}^2(x+\frac{\pi }{4}ta{n}^{2}x)=1$

• A.
• B.
• C.
• D. x=kπ, k ϵ I

Answer 1

The correct option is A

Given
$co{s}^2\left[\frac{\pi }{4}\right(sinx+\sqrt{2}co{s}^{2}x\left)\right]-ta{n}^2(x+\frac{\pi }{4}ta{n}^{2}x)=1$
$orsi{n}^2\left\{\frac{\pi }{4}(sinx+\sqrt{2}co{s}^{2}x)\right\}+ta{n}^2(x+\frac{\pi }{4}ta{n}^{2}x)=0$
It is possible only when

$si{n}^2\frac{\pi }{4}(sinx+\sqrt{2}co{s}^{2}x)=0.....\left(1\right)$
and $ta{n}^2(x+\frac{\pi }{4}ta{n}^{2}x)=0......\left(2\right)$
from equation (1) $si{n}^2\frac{\pi }{4}(sinx+\sqrt{2}co{s}^{2}x)=0$
$\therefore \frac{\pi }{4}(sinx+\sqrt{2}co{s}^{2}x)=n\pi ,n\in 1$.

or, $sinx+\sqrt{2}co{s}^{2}x=4n$
$\therefore |sinx+\sqrt{2}co{s}^{2}x|\le |sinx|+\sqrt{|}cosx{|}^2$
$\le 1+\sqrt{2}<4$
$\therefore$ The equation has no solution for $n\ne 0$ we consider n=0
$\therefore sinx+\sqrt{2}co{s}^{2}x=0$
i.e., $\sqrt{2}si{n}^{2}x-sinx-\sqrt{2}=0$
or, $(sinx-\sqrt{2})(\sqrt{2}+1=0)$
$\Theta sinx\ne \sqrt{2}\therefore sinx=-\frac{1}{\sqrt{2}}$
$\therefore$ The value of x satisfy the equation (2),

then general solution of given equation is
$x=k\pi +(-1{)}^{k+1}\frac{\pi }{4},k\in 1$