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

Question

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To solve the equation $\cos 2x = \sqrt{2} - \cos 2x$, first combine like terms to get $2\cos 2x = \sqrt{2}$. Dividing both sides by 2 gives $\cos 2x = \frac{\sqrt{2}}{2}$. The solutions to $\cos \theta = \frac{\sqrt{2}}{2}$ are at $\theta = \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4},$ and $\frac{7\pi}{4}$ within one cycle of the cosine function. However, since we have $\cos 2x$ instead of $\cos x$, we need to find the values of $x$ that satisfy $2x = \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4},$ and $\frac{7\pi}{4}$. Solving for $x$ gives $x = \frac{\pi}{8}, \frac{3\pi}{8}, \frac{5\pi}{8},$ and $\frac{7\pi}{8}$. However, the question asks for solutions in the interval $[0, 2\pi)$, and considering the periodicity of the cosine function, the correct solutions are $x = \frac{\pi}{8}, \frac{9\pi}{8}, \frac{7\pi}{8},$ and $\frac{15\pi}{8}$, which are all within the specified interval.

Solve the Equation for Solutions in the Interval [0, 2 cos⁡2x=2−cos⁡2x\cos 2 x = \sqrt { 2 } - \cos 2 xcos2x=2​−cos2x

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