If sin^2 theta - 2cos theta + frac{1}{4} = 0, | vedclass

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(B) Given the equation: $\sin^2 	heta - 2\cos 	heta + \frac{1}{4} = 0$Using the identity $\sin^2 	heta = 1 - \cos^2 	heta$,we get:$1 - \cos^2 	he

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$2n\pi \pm \frac{\pi}{3}$

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(B) Given the equation: $\sin^2 	heta - 2\cos 	heta + \frac{1}{4} = 0$Using the identity $\sin^2 	heta = 1 - \cos^2 	heta$,we get:$1 - \cos^2 	heta - 2\cos 	heta + \frac{1}{4} = 0$$-\cos^2 	heta - 2\cos 	heta + \frac{5}{4} = 0$Multiplying by $-1$:$\cos^2 	heta + 2\cos 	heta - \frac{5}{4} = 0$Using the quadratic formula $\cos 	heta = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:$\cos 	heta = \frac{-2 \pm \sqrt{4 - 4(1)(-\frac{5}{4})}}{2} = \frac{-2 \pm \sqrt{4 + 5}}{2} = \frac{-2 \pm 3}{2}$This gives two possible values:$1) \cos 	heta = \frac{-2 + 3}{2} = \frac{1}{2}$$2) \cos 	heta = \frac{-2 - 3}{2} = -\frac{5}{2}$Since $|\cos 	heta| \le 1$,the value $\cos 	heta = -\frac{5}{2}$ is rejected.Thus,$\cos 	heta = \frac{1}{2} = \cos(\frac{\pi}{3})$.The general solution for $\cos 	heta = \cos \alpha$ is $	heta = 2n\pi \pm \alpha$.Therefore,$	heta = 2n\pi \pm \frac{\pi}{3}$.

If $\sin^2 \theta - 2\cos \theta + \frac{1}{4} = 0$,then the general value of $\theta$ is

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