The general value of theta in the equation 2s | vedclass

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(C) Given the equation: $2\sqrt{3} \cos 	heta = 	an 	heta$Since $	an 	heta = \frac{\sin 	heta}{\cos 	heta}$,we have $2\sqrt{3} \cos 	heta = \f

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$n\pi + (-1)^n \frac{\pi}{3}$

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(C) Given the equation: $2\sqrt{3} \cos 	heta = 	an 	heta$Since $	an 	heta = \frac{\sin 	heta}{\cos 	heta}$,we have $2\sqrt{3} \cos 	heta = \frac{\sin 	heta}{\cos 	heta}$Multiplying both sides by $\cos 	heta$ (assuming $\cos 	heta 
eq 0$): $2\sqrt{3} \cos^2 	heta = \sin 	heta$Using $\cos^2 	heta = 1 - \sin^2 	heta$,we get $2\sqrt{3}(1 - \sin^2 	heta) = \sin 	heta$$2\sqrt{3} - 2\sqrt{3} \sin^2 	heta = \sin 	heta$$2\sqrt{3} \sin^2 	heta + \sin 	heta - 2\sqrt{3} = 0$This is a quadratic equation in $\sin 	heta$. Using the quadratic formula $\sin 	heta = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:$\sin 	heta = \frac{-1 \pm \sqrt{1^2 - 4(2\sqrt{3})(-2\sqrt{3})}}{2(2\sqrt{3})} = \frac{-1 \pm \sqrt{1 + 48}}{4\sqrt{3}} = \frac{-1 \pm 7}{4\sqrt{3}}$Case $1$: $\sin 	heta = \frac{-1 - 7}{4\sqrt{3}} = \frac{-8}{4\sqrt{3}} = -\frac{2}{\sqrt{3}} \approx -1.15$ (Impossible,as $|\sin 	heta| \leq 1$)Case $2$: $\sin 	heta = \frac{-1 + 7}{4\sqrt{3}} = \frac{6}{4\sqrt{3}} = \frac{\sqrt{3}}{2}$Since $\sin 	heta = \sin(\frac{\pi}{3})$,the general solution is $	heta = n\pi + (-1)^n \frac{\pi}{3}$.

The general value of $\theta$ in the equation $2\sqrt{3} \cos \theta = \tan \theta$ is

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