Solve the equation $3\sin \theta -\cos ^{2}\theta =3$, where $0\leq \theta <2\pi$.

Solution The equation involves both sine and cosine functions. We use the Pythagorean identity $\sin ^{2}\theta +\cos ^{2}\theta =1$ to rewrite the equation in terms of $\sin \theta .$ $$\begin{eqnarray*} 3\sin \theta -\cos ^{2}\theta &=&3 \\ 3\sin \theta -( 1-\sin ^{2}\theta)~&=&3\qquad {\color{#0066A7}{\cos ^{2}{\theta =1-}\sin ^{2}{\theta }}} \\ \sin ^{2}\theta +3\sin \theta -4 &=&0 \end{eqnarray*}$$

This is a quadratic equation in $\sin \theta$. Factor the left side and solve for $\sin \theta .$ $$\begin{array}{rcl@{\quad}c@{\quad}rcl} ( \sin \theta +4) (\sin \theta -1)~&=&0& \\ \sin \theta +4 &=&0&\hbox{ or }& \sin \theta -1&=&0 \\ \sin \theta &=&-4&\hbox{ or }& \sin \theta &=&1 \end{array}$$

The range of the sine function is $-1\leq y\leq 1,$ so $\sin \theta =-4$ has no solution. Solving $\sin \theta =1,$ we obtain $$\theta =\sin ^{-1}1=\dfrac{\pi }{2}$$

The only solution in the interval $\left[ 0,2\pi \right)$ is $\dfrac{\pi }{2 }.$