Question

$\mathrm{The}\mathrm{general}\mathrm{solution}\mathrm{of}\mathrm{the}\mathrm{equation}2{\cos }^2\theta +3\sin \theta =0\mathrm{is}$
.

• A. $n\pi +{(-1)}^n\frac{\pi }{6},n\in Z$
• B. $n\pi +{(-1)}^{n+1}\frac{\pi }{6},n\in Z$
• C. $n\pi +{(-1)}^n\frac{5\pi }{6},n\in Z$

Answer 1

The correct option is B $n\pi +{(-1)}^{n+1}\frac{\pi }{6},n\in Z$

To solve this equation, we try to write the equation in terms of $\sin \theta$.
Now, we can write the equation as:
$2{\cos }^2\theta +3\sin \theta =0\Rightarrow 2(1-{\sin }^2\theta )+3\sin \theta =0\Rightarrow 2{\sin }^2\theta -3\sin \theta -2=0\Rightarrow (\sin \theta -2)(2\sin \theta +1)=0$
This means either
$\sin \theta =2\mathrm{or}\sin \theta =\frac{-1}{2}.$
But, $\sin \theta =2$ is not possible.
Thus,
$\sin \theta =\frac{-1}{2}=\sin \left(\frac{-\pi }{6}\right)\Rightarrow \theta =n\pi +{(-1)}^n\left(\frac{-\pi }{6}\right),n\in Z\Rightarrow \theta =\mathrm{n\pi }+{(-1)}^{n+1}\left(\frac{\pi }{6}\right),n\in Z$