The principal solutions of (5+3 sin theta)(2 | vedclass

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Question

(A) Given the equation $(5+3 \sin 	heta)(2 \cos 	heta+1)=0$. This implies either $(5+3 \sin 	heta)=0$ or $(2 \cos 	heta+1)=0$. Case $1$: $5+3 \sin

Vedclass · Accepted Answer

$\frac{2 \pi}{3}, \frac{4 \pi}{3}$

Vedclass · Answer

(A) Given the equation $(5+3 \sin 	heta)(2 \cos 	heta+1)=0$. This implies either $(5+3 \sin 	heta)=0$ or $(2 \cos 	heta+1)=0$. Case $1$: $5+3 \sin 	heta = 0 \implies \sin 	heta = -\frac{5}{3}$. Since the range of $\sin 	heta$ is $[-1, 1]$,this equation has no real solution. Case $2$: $2 \cos 	heta + 1 = 0 \implies \cos 	heta = -\frac{1}{2}$. We know that $\cos 	heta = -\frac{1}{2}$ in the second and third quadrants. In the interval $[0, 2\pi)$,the solutions are $	heta = \pi - \frac{\pi}{3} = \frac{2\pi}{3}$ and $	heta = \pi + \frac{\pi}{3} = \frac{4\pi}{3}$. Thus,the principal solutions are $\frac{2\pi}{3}$ and $\frac{4\pi}{3}$.

The principal solutions of $(5+3 \sin \theta)(2 \cos \theta+1)=0$ are

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