Question

Solve in |R 2cos(2x) +√2 = 0

Answer 1

It is given that:

$\begin{gathered} 2\cos 2x+\sqrt[]{2}=0 \\ 2\cos 2x=-\sqrt[]{2} \\ \cos 2x=-\frac{\sqrt[]{2}}{2} \end{gathered}$

cos2x is negative square root 2 divided by 2 in the second and third quadrants, so it follows:

$\begin{gathered} \cos 2x=\cos \theta \\ 2x=2n\pi\pm\theta \end{gathered}$

Here theta is given by:

$\theta=\pi-\frac{\pi}{4}=\frac{3\pi}{4},\theta=\pi+\frac{\pi}{4}=\frac{5\pi}{4}$

So the solution is given by:

$\begin{gathered} 2x=2n\pi\pm\frac{3\pi}{4};2x=2n\pi\pm\frac{5\pi}{4} \\ x=n\pi\pm\frac{3\pi}{8};x=n\pi\pm\frac{5\pi}{8} \end{gathered}$