Trig general solution

**srirahulan** · August 24th 2012, 05:09 PM

If $cos^2x=\frac{1}{2}$ What is the General solutions for it?

**Prove It** · August 24th 2012, 05:22 PM

Originally Posted by srirahulan

If $cos^2x=\frac{1}{2}$ What is the General solutions for it?

$\displaystyle \begin{align*} \cos^2{x} &= \frac{1}{2} \\ \cos{x} &= \pm \frac{1}{\sqrt{2}} \\ x &= \left\{ \frac{\pi}{4}, \pi - \frac{\pi}{4}, \pi + \frac{\pi}{4}, 2\pi - \frac{\pi}{4} \right\} + 2\pi n \textrm{ where } n \in \mathbf{Z} \\ x &= \left\{ \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4} \right\} + 2\pi n \\ x &= \frac{\pi}{4} + \frac{\pi}{2} m \textrm{ where } m \in \mathbf{Z} \\ x &= \frac{\pi + 2\pi m}{4} \\ x &= \frac{( 1 + 2m )\pi}{ 4 } \end{align*}$

**srirahulan** · August 24th 2012, 05:40 PM

i get x=2npi+-pi/4 in case 1 in case 2 i get x=2npi+-3pi/4 is this correct?

**Prove It** · August 24th 2012, 05:43 PM

Originally Posted by srirahulan

i get x=2npi+-pi/4 in case 1 in case 2 i get x=2npi+-3pi/4 is this correct?

It is correct, but it is messy. What exactly was wrong with the solution I posted?

**srirahulan** · August 24th 2012, 05:46 PM

Your answer is perfectly correct.i just confirm my answer only, great thanks for you.

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