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I am trying to derive the general solution for the equation $$ 2 \cos^2 x = 1$$

It's get reduced to $$ \cos x = \pm \frac1 {\sqrt{2}} $$

Now, for, $\cos x = \frac 1 {\sqrt{2}} \Rightarrow x = 2n\pi \pm \frac{\pi}{4}$ and for $\cos x = - \frac 1 {\sqrt{2}} \Rightarrow x = 2n\pi \pm \frac{3\pi}{4}$, but I am stuck here I want to know how to derive the general solution that will satisfy the mother equation ($ 2 \cos^2 x = 1$) itself?

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    You've found all the solutions already, so what's the problem? Anyway, you can also solve the equation by rewriting it as $\cos 2x = 0$.2012-04-08

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Looking at the unit circle, we see that odd integer multiples of $\frac{\pi}{4}$ give all solutions.

If you wish to arrive at this from the equations that you derived, notice that for a given $n$, the solutions corresponding to " $...+\pi /4$ " and " $...-3 \pi /4 $" are pi units apart.

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    That's a good point! I tried to think on those lines and I think $\frac{n\pi}{2} \pm \frac{\pi}4 , \forall n \in \mathbb{W}$ gives all the solutions too isn't?2012-04-08
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    Odd integer multiples --> $$(2n \pm 1) \dfrac{\pi}{4} = (2n) \dfrac{\pi}{4} \pm (1) \dfrac{\pi}{4} = \dfrac{n \pi}{2} \pm \dfrac{\pi}{4}$$2012-04-08
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    Thanks a lot Chaz!! Actually I did the whole thing reverse and also thanks for helping me understand how to get it from my derivation too:)2012-04-08