Solve the equation for solutions in the interval [0, 2 π).
-$$\sqrt { 2 } \cos 2 x = 1$$

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Quizplus · Accepted Answer

First, solve the equation $\sqrt{2} \cos 2x = 1$ by dividing both sides by $\sqrt{2}$, giving $\cos 2x = \frac{1}{\sqrt{2}}$. This simplifies to $\cos 2x = \cos \frac{\pi}{4}$, because $\cos \frac{\pi}{4} = \frac{1}{\sqrt{2}}$. The solutions to $\cos 2x = \cos \frac{\pi}{4}$ in the interval $[0, 2\pi)$ are $x = \frac{\pi}{8}, \frac{9\pi}{8}, \frac{7\pi}{8}, \frac{15\pi}{8}$, since $\cos \theta = \cos \phi$ has solutions $\theta = \phi + 2k\pi$ and $\theta = -\phi + 2k\pi$, where $k$ is an integer.

Solve the Equation for Solutions in the Interval [0, 2 2cos⁡2x=1\sqrt { 2 } \cos 2 x = 12​cos2x=1

Solve the Equation for Solutions in the Interval [0, 2 $\sqrt { 2 } \cos 2 x = 1$