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-(\pi)/(4)=(3\pi)/(4),\theta=\pi+(\pi)/(4)=(5\pi)/(4)`

So the solution is given by:

`\begin{gathered} 2x=2n\pi\pm(3\pi)/(4);2x=2n\pi\pm(5\pi)/(4) \\ x=n\pi\pm(3\pi)/(8);x=n\pi\pm(5\pi)/(8) \end{gathered}`