To solve: the equation \cos \theta (2 \sin \theta + 1)=0

Approach:
The domain of the trigonometry function of ${\displaystyle \cos \theta }$ is lies between ${\displaystyle [-1,1]}$. No solution exists beyond this domain.
Obtain factor of given equation and simply the factor equations to obtain solutions.
Cosine and sine have period ${\displaystyle 2\pi }$, thus find the solution in any interval of length ${\displaystyle 2\pi }$.
Sine function is positive in first and second quadrant. Cosine function is positive in first and fourth quadrant.
Calculation:
Consider the trigonometry equation.
${\displaystyle \cos \theta (2\sin \theta +1)=0}$
The factors of above equation are,
${\displaystyle 2\sin \theta +1=0}$
${\displaystyle \cos \theta =0}$
The solution obtained for the factor in which sine and cosine functions are involved so we will get the solution in the interval of ${\displaystyle [0,2\pi ]}$.
Consider the factors.
${\displaystyle 2\sin \theta +1=0\dots \left(1\right)}$
Substract 1 both sides in equation (1).
${\displaystyle 2\sin \theta =-1}$
Divide by 2 both sides in equation (1).
${\displaystyle 2\sin \theta =-1}$
${\displaystyle \sin \theta =-\frac{1}{2}}$
Multiply by ${\displaystyle {\sin }^{-1}}$ both sides in equation (1).
${\displaystyle {\sin }^{-1}\sin \theta ={\sin }^{-1}(-\frac{1}{2})}$
${\displaystyle \theta ={\sin }^{-1}(-\frac{1}{2})}$
${\displaystyle \theta =\frac{7\pi }{6},\frac{11\pi }{6}}$
The sine has period, ${\displaystyle 2\pi }$, so we get all solution of the equation by adding integer multiples of ${\displaystyle 2\pi }$ to these solutions:
${\displaystyle \theta =\frac{7\pi }{6}+2k\pi }$
${\displaystyle \theta =\frac{11\pi }{6}+2k\pi }$
and,
Consider the factors.
${\displaystyle \cos \theta =0}$
${\displaystyle \theta =\frac{\pi }{2},\frac{3\pi }{2}}$
The solution repeats value of the equation at every length of ${\displaystyle \pi }$ in the interval ${\displaystyle [0,2\pi ]}$.
We will get all solutions of the equation by adding integer multiples of ${\displaystyle \pi }$ to these solutions:
${\displaystyle \theta =\frac{\pi }{2}+k\pi }$
Therefore, the solution of the trigonometry equation ${\displaystyle \cos \theta (2\sin \theta +1)=0}$ are ${\displaystyle \theta =\frac{\pi }{2}+k\pi ,\theta =\frac{7\pi }{6}+2k\pi }$ and ${\displaystyle \theta =\frac{11\pi }{6}+2k\pi }$.
Conclusion:
Hence, the solutions of the trigonometry equation ${\displaystyle \cos \theta (2\sin \theta +1)=0}$ are ${\displaystyle \theta =\frac{\pi }{2}+k\pi ,\theta =\frac{7\pi }{6}+2k\pi }$ and ${\displaystyle \theta =\frac{11\pi }{6}+2k\pi }$