\(2\sin\theta\cos\theta+\cos\theta\ =\ 0\)
Factor cos θ out of both terms.
\((\cos\theta)(2\sin\theta+1)\ =\ 0\)
Set each factor equal to 0 and solve for θ .
And remember that we only want solutions for θ in the interval [0, 2π]
\(\begin{array}{} \cos\theta\ =\ 0&\qquad\text{or}\qquad&2\sin\theta+1\ =\ 0\\~\\ \theta\ =\ \frac{\pi}{2}&&2\sin\theta\ =\ -1\\~\\ \theta\ =\ \frac{3\pi}{2}&&\sin\theta\ =\ -\frac{1}{2}\\~\\ &&\theta\ =\ \frac{7\pi}{6}\\~\\ &&\theta\ =\ \frac{11\pi}{6} \end{array}\)
So the values for θ in the interval [0, 2π] that satisfy the equation are:
\(\frac{\pi}{2},\ \frac{3\pi}{2},\ \frac{7\pi}{6},\ \frac{11\pi}{6}\)_
