The set of angles between 0 and 2pi satisfyin | vedclass

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(B) Given equation: $4\cos^2 	heta - 2\sqrt{2}\cos 	heta - 1 = 0$. Using the quadratic formula for $\cos 	heta$: $\cos 	heta = \frac{2\sqrt{2} \pm

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$\left\{ \frac{\pi}{12}, \frac{7\pi}{12}, \frac{17\pi}{12}, \frac{23\pi}{12} \right\}$

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(B) Given equation: $4\cos^2 	heta - 2\sqrt{2}\cos 	heta - 1 = 0$. Using the quadratic formula for $\cos 	heta$: $\cos 	heta = \frac{2\sqrt{2} \pm \sqrt{(2\sqrt{2})^2 - 4(4)(-1)}}{2(4)} = \frac{2\sqrt{2} \pm \sqrt{8 + 16}}{8} = \frac{2\sqrt{2} \pm \sqrt{24}}{8} = \frac{2\sqrt{2} \pm 2\sqrt{6}}{8} = \frac{\sqrt{2} \pm \sqrt{6}}{4}$. Case $1$: $\cos 	heta = \frac{\sqrt{6} + \sqrt{2}}{4} = \cos\left(\frac{\pi}{12}ight)$. Thus,$	heta = \frac{\pi}{12}$ or $	heta = 2\pi - \frac{\pi}{12} = \frac{23\pi}{12}$. Case $2$: $\cos 	heta = \frac{\sqrt{2} - \sqrt{6}}{4} = -\left(\frac{\sqrt{6} - \sqrt{2}}{4}ight) = -\cos\left(\frac{\pi}{12}ight) = \cos\left(\pi - \frac{\pi}{12}ight) = \cos\left(\frac{11\pi}{12}ight)$. Wait,checking the value: $\cos\left(\frac{7\pi}{12}ight) = \cos\left(\frac{\pi}{3} + \frac{\pi}{4}ight) = \cos\frac{\pi}{3}\cos\frac{\pi}{4} - \sin\frac{\pi}{3}\sin\frac{\pi}{4} = \frac{1}{2}\cdot\frac{\sqrt{2}}{2} - \frac{\sqrt{3}}{2}\cdot\frac{\sqrt{2}}{2} = \frac{\sqrt{2}-\sqrt{6}}{4}$. So,$\cos 	heta = \cos\left(\frac{7\pi}{12}ight)$. Thus,$	heta = \frac{7\pi}{12}$ or $	heta = 2\pi - \frac{7\pi}{12} = \frac{17\pi}{12}$. The set of solutions is $\left\{ \frac{\pi}{12}, \frac{7\pi}{12}, \frac{17\pi}{12}, \frac{23\pi}{12} ight\}$.

The set of angles between $0$ and $2\pi$ satisfying the equation $4\cos^2 \theta - 2\sqrt{2}\cos \theta - 1 = 0$ is

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