 To solve: the equation \cos \theta (2 \sin \theta + 1)=0 Nannie Mack 2021-08-11 Answered
To solve:
The equation $$\displaystyle{\cos{\theta}}{\left({2}{\sin{\theta}}+{1}\right)}={0}$$

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Approach:
The domain of the trigonometry function of $$\displaystyle{\cos{\theta}}$$ is lies between $$\displaystyle{\left[-{1},{1}\right]}$$. 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.
$$\displaystyle{\cos{\theta}}{\left({2}{\sin{\theta}}+{1}\right)}={0}$$
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 $$\displaystyle{\left[{0},{2}\pi\right]}$$.
Consider the factors.
$$\displaystyle{2}{\sin{\theta}}+{1}={0}\ldots{\left({1}\right)}$$
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).
$$\displaystyle{{\sin}^{{-{1}}}{\sin{\theta}}}={{\sin}^{{-{1}}}{\left(-{\frac{{{1}}}{{{2}}}}\right)}}$$
$$\displaystyle\theta={{\sin}^{{-{1}}}{\left(-{\frac{{{1}}}{{{2}}}}\right)}}$$
$$\displaystyle\theta={\frac{{{7}\pi}}{{{6}}}},{\frac{{{11}\pi}}{{{6}}}}$$
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:
$$\displaystyle\theta={\frac{{{7}\pi}}{{{6}}}}+{2}{k}\pi$$
$$\displaystyle\theta={\frac{{{11}\pi}}{{{6}}}}+{2}{k}\pi$$
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 $$\displaystyle{\left[{0},{2}\pi\right]}$$.
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 $$\displaystyle{\cos{\theta}}{\left({2}{\sin{\theta}}+{1}\right)}={0}$$ are $$\displaystyle\theta={\frac{{\pi}}{{{2}}}}+{k}\pi,\theta={\frac{{{7}\pi}}{{{6}}}}+{2}{k}\pi$$ and $$\displaystyle\theta={\frac{{{11}\pi}}{{{6}}}}+{2}{k}\pi$$.
Conclusion:
Hence, the solutions of the trigonometry equation $$\displaystyle{\cos{\theta}}{\left({2}{\sin{\theta}}+{1}\right)}={0}$$ are $$\displaystyle\theta={\frac{{\pi}}{{{2}}}}+{k}\pi,\theta={\frac{{{7}\pi}}{{{6}}}}+{2}{k}\pi$$ and $$\displaystyle\theta={\frac{{{11}\pi}}{{{6}}}}+{2}{k}\pi$$