If 4sin^2 theta + 2(sqrt{3} + 1)cos theta = 4 | vedclass

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Question

(A) Given equation: $4\sin^2 	heta + 2(\sqrt{3} + 1)\cos 	heta = 4 + \sqrt{3}$ Substitute $\sin^2 	heta = 1 - \cos^2 	heta$: $4(1 - \cos^2 	heta)

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$2n\pi \pm \frac{\pi}{6}$ or $2n\pi \pm \frac{\pi}{3}$

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(A) Given equation: $4\sin^2 	heta + 2(\sqrt{3} + 1)\cos 	heta = 4 + \sqrt{3}$ Substitute $\sin^2 	heta = 1 - \cos^2 	heta$: $4(1 - \cos^2 	heta) + 2(\sqrt{3} + 1)\cos 	heta = 4 + \sqrt{3}$ $4 - 4\cos^2 	heta + 2(\sqrt{3} + 1)\cos 	heta = 4 + \sqrt{3}$ $4\cos^2 	heta - 2(\sqrt{3} + 1)\cos 	heta + \sqrt{3} = 0$ Let $x = \cos 	heta$. Then $4x^2 - 2(\sqrt{3} + 1)x + \sqrt{3} = 0$ $4x^2 - 2\sqrt{3}x - 2x + \sqrt{3} = 0$ $2x(2x - \sqrt{3}) - 1(2x - \sqrt{3}) = 0$ $(2x - 1)(2x - \sqrt{3}) = 0$ So,$x = \frac{1}{2}$ or $x = \frac{\sqrt{3}}{2}$ Case $1$: $\cos 	heta = \frac{1}{2} = \cos(\frac{\pi}{3}) \Rightarrow 	heta = 2n\pi \pm \frac{\pi}{3}$ Case $2$: $\cos 	heta = \frac{\sqrt{3}}{2} = \cos(\frac{\pi}{6}) \Rightarrow 	heta = 2n\pi \pm \frac{\pi}{6}$ Thus,the general solution is $	heta = 2n\pi \pm \frac{\pi}{6}$ or $2n\pi \pm \frac{\pi}{3}$.

If $4\sin^2 \theta + 2(\sqrt{3} + 1)\cos \theta = 4 + \sqrt{3}$,then the general value of $\theta$ is

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