If theta in [0, 2pi] and cos 2theta = cos the | vedclass

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(C) Given,$\cos 2	heta = \cos 	heta + \sin 	heta$. Using the identity $\cos 2	heta = \cos^2 	heta - \sin^2 	heta$,we get: $\cos^2 	heta - \sin^

Vedclass · Accepted Answer

$\frac{24\pi}{4}$

Vedclass · Answer

(C) Given,$\cos 2	heta = \cos 	heta + \sin 	heta$. Using the identity $\cos 2	heta = \cos^2 	heta - \sin^2 	heta$,we get: $\cos^2 	heta - \sin^2 	heta = \cos 	heta + \sin 	heta$ $(\cos 	heta + \sin 	heta)(\cos 	heta - \sin 	heta) = \cos 	heta + \sin 	heta$ $(\cos 	heta + \sin 	heta)(\cos 	heta - \sin 	heta - 1) = 0$ This implies $\cos 	heta + \sin 	heta = 0$ or $\cos 	heta - \sin 	heta = 1$. Case $1$: $\cos 	heta + \sin 	heta = 0 \Rightarrow 	an 	heta = -1$. For $	heta \in [0, 2\pi]$,$	heta = \frac{3\pi}{4}, \frac{7\pi}{4}$. Case $2$: $\cos 	heta - \sin 	heta = 1$. Dividing by $\sqrt{2}$,we get $\frac{1}{\sqrt{2}}\cos 	heta - \frac{1}{\sqrt{2}}\sin 	heta = \frac{1}{\sqrt{2}}$ $\Rightarrow \cos(	heta + \frac{\pi}{4}) = \cos(\frac{\pi}{4})$. Thus,$	heta + \frac{\pi}{4} = 2n\pi \pm \frac{\pi}{4}$. For $n=0$,$	heta = 0$ or $	heta = -\frac{\pi}{2} \equiv \frac{3\pi}{2}$. For $n=1$,$	heta = 2\pi$ or $	heta = \pi$. Checking $	heta = \pi$ in the original equation: $\cos(2\pi) = 1$,$\cos \pi + \sin \pi = -1 + 0 = -1$. $1 
eq -1$,so $\pi$ is not a solution. The valid values are $	heta \in \{0, \frac{3\pi}{4}, \frac{3\pi}{2}, \frac{7\pi}{4}, 2\pi\}$. Sum $= 0 + \frac{3\pi}{4} + \frac{6\pi}{4} + \frac{7\pi}{4} + \frac{8\pi}{4} = \frac{24\pi}{4} = 6\pi$.

If $\theta \in [0, 2\pi]$ and $\cos 2\theta = \cos \theta + \sin \theta$,then the sum of all values of $\theta$ satisfying the equation is

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