Find the common solution of the equations -
$$2\sin^{2}x+\sin^{2}2x=2 $$and$$\sin2x+\cos2x=tanx $$

Question

Find the common solution of the equations -
$$2\sin^{2}x+\sin^{2}2x=2 $$and$$\sin2x+\cos2x=tanx $$

anonymous · Answer

play around first one with$$\sin^2 2x=4\sin^2 x\cos^2x$$and replace $\cos^2x$  with  $1-\sin^2 x$

anonymous · Answer

For 1st eq.$$x=n \pi \pm \frac{\pi}{4}$$
or$$x=n \pi \pm \frac{\pi}{2}$$

anonymous · Answer

For 2nd eq.
$$x=\frac{n \pi}{2}-(-1)^{n}\frac{\pi}{4}$$
or$$x={n \pi}{}\pm\frac{\pi}{4}$$

anonymous · Answer

@mukushla Now what to do??

experimentx · Answer

if you do that ... the first equation is quadratic in sin^2 x

anonymous · Answer

after solving both quadratics i got the above sol.

experimentx · Answer

anonymous · Answer

ans. common sol.- $$x=(2n+1)\frac{\pi}{4}$$

anonymous · Answer

????

experimentx · Answer

anonymous · Answer

i dont understand that solution

experimentx · Answer

how did you solve the second equation?

anonymous · Answer

$$2sinxcosx+1-2\sin^{2}x=\frac{sinx}{cosx}$$

anonymous · Answer

$$2sinxcos^{2}x+cosx-2\sin^{2}xcosx-{sinx}=0$$

anonymous · Answer

$$(2sinxcos+1)(cosx-sinx)=0$$

experimentx · Answer

anonymous · Answer

Can't we use tan formula there ??

anonymous · Answer

how?

anonymous · Answer

$$\frac{2tanx}{1 + \tan^2x} + \frac{1 - \tan^2x}{1 + \tan^2x} = tanx$$

anonymous · Answer

Or may be I am wrong too or may be I am going long by using that method..

experimentx · Answer

experimentx · Answer

$$ \sqrt 2 \sin\left( 2x + {\pi \over 4}\right) = \tan x$$
the solution seems correct. maybe i'm using WA too much for free.

experimentx · Answer

Damn ... wolf was right ... sqrt(2) arctan(1 -sqrt(2)) = pi/4 (didn't know this identity) http://www.wolframalpha.com/input/?i=2+arctan%281+-+sqrt%282%29%29