The common roots of the equations 2sin^2 x + | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Given equations are:$2\sin^2 x + \sin^2 2x = 2$ ... $(i)$$\sin 2x + \cos 2x = 	an x$ ... $(ii)$From $(i)$:$2\sin^2 x + (2\sin x \cos x)^2 = 2$$2\

Vedclass · Accepted Answer

$x = (2n + 1)\frac{\pi}{4}$

Vedclass · Answer

(B) Given equations are:$2\sin^2 x + \sin^2 2x = 2$ ... $(i)$$\sin 2x + \cos 2x = 	an x$ ... $(ii)$From $(i)$:$2\sin^2 x + (2\sin x \cos x)^2 = 2$$2\sin^2 x + 4\sin^2 x \cos^2 x = 2$$2\sin^2 x + 4\sin^2 x (1 - \sin^2 x) = 2$$2\sin^2 x + 4\sin^2 x - 4\sin^4 x = 2$$4\sin^4 x - 6\sin^2 x + 2 = 0$$2\sin^4 x - 3\sin^2 x + 1 = 0$$(2\sin^2 x - 1)(\sin^2 x - 1) = 0$So,$\sin^2 x = \frac{1}{2}$ or $\sin^2 x = 1$.If $\sin^2 x = 1$,then $\cos^2 x = 0$,which makes $	an x$ undefined.If $\sin^2 x = \frac{1}{2}$,then $\cos^2 x = \frac{1}{2}$,so $	an^2 x = 1$,which means $x = n\pi \pm \frac{\pi}{4}$.From $(ii)$:$\frac{2	an x}{1 + 	an^2 x} + \frac{1 - 	an^2 x}{1 + 	an^2 x} = 	an x$$2	an x + 1 - 	an^2 x = 	an x + 	an^3 x$$	an^3 x + 	an^2 x - 	an x - 1 = 0$$	an^2 x(	an x + 1) - 1(	an x + 1) = 0$$(	an^2 x - 1)(	an x + 1) = 0$So,$	an^2 x = 1$ or $	an x = -1$.Both imply $	an x = \pm 1$,which gives $x = n\pi \pm \frac{\pi}{4} = (2n + 1)\frac{\pi}{4}$.Thus,the common roots are $x = (2n + 1)\frac{\pi}{4}$.

The common roots of the equations $2\sin^2 x + \sin^2 2x = 2$ and $\sin 2x + \cos 2x = \tan x$ are

Similar Questions

In $\Delta ABC,$ if $\sin^2 \frac{A}{2}, \sin^2 \frac{B}{2}, \sin^2 \frac{C}{2}$ are in $H.P.,$ then $a, b, c$ will be in

The set of values of $a$ for which the equation $cos\, 2x + a\, sin\, x = 2a - 7$ possesses a solution is:

The smallest positive root of the equation $\tan x - x = 0$ lies in the interval

In any triangle $ABC$,${\sin ^2}\frac{A}{2} + {\sin ^2}\frac{B}{2} + {\sin ^2}\frac{C}{2}$ is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module