If cos 2theta + 3cos theta = 0,then the gener | vedclass

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(A) Given the equation $\cos 2	heta + 3\cos 	heta = 0$.Using the identity $\cos 2	heta = 2\cos^2 	heta - 1$,we get:$2\cos^2 	heta - 1 + 3\cos 	h

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$2n\pi \pm \cos^{-1}\left(\frac{-3 + \sqrt{17}}{4}\right)$

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(A) Given the equation $\cos 2	heta + 3\cos 	heta = 0$.Using the identity $\cos 2	heta = 2\cos^2 	heta - 1$,we get:$2\cos^2 	heta - 1 + 3\cos 	heta = 0$$2\cos^2 	heta + 3\cos 	heta - 1 = 0$Using the quadratic formula for $\cos 	heta$:$\cos 	heta = \frac{-3 \pm \sqrt{3^2 - 4(2)(-1)}}{2(2)} = \frac{-3 \pm \sqrt{9 + 8}}{4} = \frac{-3 \pm \sqrt{17}}{4}$Since $-1 \le \cos 	heta \le 1$,we check the values:$\frac{-3 - \sqrt{17}}{4} \approx \frac{-3 - 4.12}{4} \approx -1.78$ (which is less than $-1$,so this is rejected).$\frac{-3 + \sqrt{17}}{4} \approx \frac{-3 + 4.12}{4} \approx 0.28$ (which is valid).Thus,$\cos 	heta = \frac{-3 + \sqrt{17}}{4}$.The general solution for $\cos 	heta = \alpha$ is $	heta = 2n\pi \pm \cos^{-1}(\alpha)$.Therefore,$	heta = 2n\pi \pm \cos^{-1}\left(\frac{-3 + \sqrt{17}}{4}ight)$.

If $\cos 2\theta + 3\cos \theta = 0$,then the general value of $\theta$ is

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