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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Expert Answer

okomgcae
Answered 2021-08-12 Author has 13765 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.
\(\displaystyle{\cos{\theta}}{\left({2}{\sin{\theta}}+{1}\right)}={0}\)
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 \(\displaystyle\theta={\frac{{{11}\pi}}{{{6}}}}+{2}{k}\pi\)
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