Question

Find the general solution of qrt%7B2%7D" id="TexFormula1" title="2cos(2x + \frac{\pi }{4} ) = \sqrt{2}" alt="2cos(2x + \frac{\pi }{4} ) = \sqrt{2}" align="absmiddle" class="latex-formula"> =

Answer 1

Answer:

$\large\boxed{x=-\dfrac{\pi}{4}+k\pi\ \text{or}\ x=k\pi\ \text{for}\ k\in\mathbb{Z}}$

Step-by-step explanation:

$2\cos\left(2x+\dfrac{\pi}{4}\right)=\sqrt2\qquad\text{divide both sides by 2}\\\\\cos\left(2x+\dfrac{\pi}{4}\right)=\dfrac{\sqrt2}{2}\to2x+\dfrac{\pi}{4}=\pm\dfrac{\pi}{4}+2k\pi\\\\2x+\dfrac{\pi}{4}=-\dfrac{\pi}{4}+2k\pi\ \vee\ 2x+\dfrac{\pi}{4}=\dfrac{\pi}{4}+2k\pi\qquad\text{subtract}\ \dfrac{\pi}{4}\ \text{from all sides}\\\\2x=-\dfrac{2\pi}{4}+2k\pi\ \vee\ 2x=2k\pi\qquad\text{divide all sides by 2}\\\\x=-\dfrac{\pi}{4}+k\pi\ \vee\ x=k\pi\ \text{for}\ k\in\mathbb{Z}$