# find the values of x

I am trying to understand how to do this problem. Can someone guide me please?

Here is the problem:

cos^2 (2x) +2sin^2(2x)=2

Here is what I have tried:
(cos2x)^2+2(sin2x)^2=2

Then,
cos2x=1-2sin^2 x
sin2x=2sinxcosx

(1-2sin^2 x)^2 + 2(2sinxcosx)^2=2
(1-4sin^2 x+ 4sin^4x+8sin^2x(1-sin^2 x)
1-sin^2 x +4sin^4 x + 8sin^2 x -8sin^4 x=2
1-7sin^2x-4sin^4x=2
-4sin^4x-7sin^2x+1-2=0
-4sin^4x -7sin^2x-1=0


then, I am either stuck or I have a wrong solution.

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Hint: Remember $\cos^2 2x + \sin^2 2x = 1$. Now look how that differs from your problem equation. – Macavity Sep 17 '13 at 3:50
Sorry I did not get what you said... – eLg Sep 17 '13 at 3:51
Anything more seems solving it completely for you - suggest you take a bit more time to think on it. – Macavity Sep 17 '13 at 3:55

I will expand on the hint, and let you finish the problem.

$$\cos^2(2x) + 2\sin^2(2x) = 2$$

What Macavity is saying is that for any real number $u$ we know that $$sin^2(u) + cos^2(u) = 1$$ So in this case, just think of $u$ as $2x$ then we have

$$\cos^2(2x) + 2\sin^2(2x) = (\cos^2(2x) + \sin^2(2x)) + \sin^2(2x) = 1 + \sin^2(2x)$$

So that $$1 + \sin^2(2x) = 2$$

Try and go on from there, and if you need more help, I can edit the answer.

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