Solving a Trigonometric Equation

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 \(\dfrac{\pi }{2 }.\)