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Find the general solution of
2cos(2x + (\pi )/(4) ) = √(2) =

1 Answer

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Answer:


\large\boxed{x=-(\pi)/(4)+k\pi\ \text{or}\ x=k\pi\ \text{for}\ k\in\mathbb{Z}}

Explanation:


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

Find the general solution of 2cos(2x + (\pi )/(4) ) = √(2) =-example-1
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