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.

\(\displaystyle{4}{\cos{\theta}}{\sin{\theta}}+{3}{\cos{\theta}}={0}\)

The factors of above equation are obtained by,

\(\displaystyle{4}{\cos{\theta}}{\sin{\theta}}+{3}{\cos{\theta}}={0}\)

\(\displaystyle{\cos{\theta}}{\left({4}{\sin{\theta}}+{3}\right)}={0}\)

The factors are,

\(\displaystyle{4}{\sin{\theta}}+{3}={0}\ldots{\left({1}\right)}\)

\(\displaystyle{\cos{\theta}}={0}\ldots{\left({2}\right)}\)

The solution obtained for the factor in which sine and cosine functions are involved so we will get the solution in the interval of \(\displaystyle{\left[{0},{2}\pi\right]}\)

Substract 3 both sides in equation (1).

\(\displaystyle{4}{\sin{\theta}}=-{3}\)

Divide by 4 both sides in equation (1).

\(\displaystyle{\sin{\theta}}=-{\frac{{{3}}}{{{4}}}}\)

Multiple by \(\displaystyle{{\sin}^{{-{1}}}}\) both sides in equation (1).

\(\displaystyle{{\sin}^{{-{1}}}{\sin{\theta}}}={{\sin}^{{-{1}}}{\left(-{\frac{{{3}}}{{{4}}}}\right)}}\)

\(\displaystyle\theta={{\sin}^{{-{1}}}{\left(-{\frac{{{3}}}{{{4}}}}\right)}}\)

\(\displaystyle={4},{5.435}\)

Here solutions are in radian.

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:

\(\displaystyle\theta={4}+{2}{k}\pi\)

\(\displaystyle\theta={5.435}+{2}{k}\pi\)

Consider the factor in equation (2).

\(\displaystyle{\cos{\theta}}={0}\)

\(\displaystyle\theta={\frac{{\pi}}{{{2}}}},{\frac{{{3}\pi}}{{{2}}}}\)

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 solutions of the trigonometry equation \(\displaystyle{4}{\cos{\theta}}{\sin{\theta}}+{3}{\cos{\theta}}={0}\) are \(\displaystyle\theta={\frac{{\pi}}{{{2}}}}+{k}\pi,\theta={4}+{2}{k}\pi\) and \(\displaystyle\theta={5.435}+{2}{k}\pi\)

Conclusion:

Hence, the solutions of the trigonometry equation \(\displaystyle{4}{\cos{\theta}}{\sin{\theta}}+{3}{\cos{\theta}}={0}\) are \(\displaystyle\theta={\frac{{\pi}}{{{2}}}}+{k}\pi,\theta={4}+{2}{k}\pi\) and \(\displaystyle\theta={5.435}+{2}{k}\pi\)

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.

\(\displaystyle{4}{\cos{\theta}}{\sin{\theta}}+{3}{\cos{\theta}}={0}\)

The factors of above equation are obtained by,

\(\displaystyle{4}{\cos{\theta}}{\sin{\theta}}+{3}{\cos{\theta}}={0}\)

\(\displaystyle{\cos{\theta}}{\left({4}{\sin{\theta}}+{3}\right)}={0}\)

The factors are,

\(\displaystyle{4}{\sin{\theta}}+{3}={0}\ldots{\left({1}\right)}\)

\(\displaystyle{\cos{\theta}}={0}\ldots{\left({2}\right)}\)

The solution obtained for the factor in which sine and cosine functions are involved so we will get the solution in the interval of \(\displaystyle{\left[{0},{2}\pi\right]}\)

Substract 3 both sides in equation (1).

\(\displaystyle{4}{\sin{\theta}}=-{3}\)

Divide by 4 both sides in equation (1).

\(\displaystyle{\sin{\theta}}=-{\frac{{{3}}}{{{4}}}}\)

Multiple by \(\displaystyle{{\sin}^{{-{1}}}}\) both sides in equation (1).

\(\displaystyle{{\sin}^{{-{1}}}{\sin{\theta}}}={{\sin}^{{-{1}}}{\left(-{\frac{{{3}}}{{{4}}}}\right)}}\)

\(\displaystyle\theta={{\sin}^{{-{1}}}{\left(-{\frac{{{3}}}{{{4}}}}\right)}}\)

\(\displaystyle={4},{5.435}\)

Here solutions are in radian.

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:

\(\displaystyle\theta={4}+{2}{k}\pi\)

\(\displaystyle\theta={5.435}+{2}{k}\pi\)

Consider the factor in equation (2).

\(\displaystyle{\cos{\theta}}={0}\)

\(\displaystyle\theta={\frac{{\pi}}{{{2}}}},{\frac{{{3}\pi}}{{{2}}}}\)

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 solutions of the trigonometry equation \(\displaystyle{4}{\cos{\theta}}{\sin{\theta}}+{3}{\cos{\theta}}={0}\) are \(\displaystyle\theta={\frac{{\pi}}{{{2}}}}+{k}\pi,\theta={4}+{2}{k}\pi\) and \(\displaystyle\theta={5.435}+{2}{k}\pi\)

Conclusion:

Hence, the solutions of the trigonometry equation \(\displaystyle{4}{\cos{\theta}}{\sin{\theta}}+{3}{\cos{\theta}}={0}\) are \(\displaystyle\theta={\frac{{\pi}}{{{2}}}}+{k}\pi,\theta={4}+{2}{k}\pi\) and \(\displaystyle\theta={5.435}+{2}{k}\pi\)