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(B) Given the equation: $2\sin 	heta + 	an 	heta = 0$We can write $	an 	heta$ as $\frac{\sin 	heta}{\cos 	heta}$:$2\sin 	heta + \frac{\sin 	h

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

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(B) Given the equation: $2\sin 	heta + 	an 	heta = 0$We can write $	an 	heta$ as $\frac{\sin 	heta}{\cos 	heta}$:$2\sin 	heta + \frac{\sin 	heta}{\cos 	heta} = 0$Factor out $\sin 	heta$:$\sin 	heta \left( 2 + \frac{1}{\cos 	heta} ight) = 0$This gives two cases:Case $1$: $\sin 	heta = 0 \Rightarrow 	heta = n\pi$,where $n \in \mathbb{Z}$.Case $2$: $2 + \frac{1}{\cos 	heta} = 0$ $\Rightarrow \frac{1}{\cos 	heta} = -2$ $\Rightarrow \cos 	heta = -\frac{1}{2}$.Since $\cos 	heta = -\frac{1}{2} = \cos \left( \frac{2\pi}{3} ight)$,the general solution is $	heta = 2n\pi \pm \frac{2\pi}{3}$,where $n \in \mathbb{Z}$.Combining both cases,the general values of $	heta$ are $n\pi$ and $2n\pi \pm \frac{2\pi}{3}$.

If $2\sin \theta + \tan \theta = 0$,then the general values of $\theta$ are

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