To solve: The equation 4 \cos \theta \sin \theta + 3 \cos \theta = 0

BolkowN 2021-07-31 Answered
To solve:
The equation \(\displaystyle{4}{\cos{\theta}}{\sin{\theta}}+{3}{\cos{\theta}}={0}\)

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Expert Answer

Willie
Answered 2021-08-01 Author has 26956 answers
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 \(\displaystyle\theta={5.435}+{2}{k}\pi\)
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