Solve the following equation \[2{{\sin }^{2}}\theta =3\cos \theta \], \[0\le \theta \le 2\pi \].

Hint: In the above equation we will use the trigonometric identity which given as follows:
\[{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1\]
After using the identity, we will solve the quadratic equation in \[\cos \theta \] and we will get the solution for the equation.

Complete Step-by-step answer:
We have been given the equation is \[2{{\sin }^{2}}\theta =3\cos \theta \].
As we know that \[{{\sin }^{2}}\theta =1-{{\cos }^{2}}\theta \].
So by using this in the above equation, we get as follows:
\[\begin{align}
  & 2\left( 1-{{\cos }^{2}}\theta \right)=3\cos \theta \\
 & 2-2{{\cos }^{2}}\theta =3\cos \theta \\
\end{align}\]
Take all the terms of left hand side of the equation to right hand side, we get as follows:
\[\begin{align}
  & 0=3\cos \theta +2{{\cos }^{2}}\theta -2 \\
 & 2{{\cos }^{2}}\theta +3\cos \theta -2 \\
\end{align}\]
Now we can write \[3\cos \theta \] in the form of \[\left( 4\cos \theta -\cos \theta \right)\] in order to factorize the above quadratic equation in ‘\[\cos \theta \]’.
\[\begin{align}
  & 2{{\cos }^{2}}\theta +4\cos \theta -\cos \theta -2=0 \\
 & 2\cos \theta \left( \cos \theta +2 \right)-1\left( \cos \theta +2 \right)=0 \\
\end{align}\]
After taking \[\left( \cos \theta +2 \right)\] as common from the above equation, we get as follows:
\[\left( \cos \theta +2 \right)\left( 2\cos \theta -1 \right)=0\]
So \[\Rightarrow \cos \theta +2=0\] and \[2\cos \theta -1=0\].
As we know the range of \[\cos \theta \] is \[\left[ -1,1 \right]\].
\[\Rightarrow \cos \theta \ne -2\]
So we have only \[2\cos \theta -1=0\].
\[\Rightarrow \cos \theta =\dfrac{1}{2}\]
As we know that the general solution of \[\cos \theta =\cos \alpha \] is given by \[\theta =2n\pi \pm \alpha \] where ‘n’ is any integer.
\[\begin{align}
  & \cos \theta =\cos \dfrac{\pi }{3} \\
 & \theta =2n\pi \pm \dfrac{\pi }{3} \\
\end{align}\]
As we have been given that \[0\le \theta \le 2\pi \].
\[\Rightarrow \theta =\dfrac{\pi }{3},\dfrac{5\pi }{3}\]
Therefore, the solution for the given equations are \[\theta =\dfrac{\pi }{3}\] and \[\theta =\dfrac{5\pi }{3}\].

Note: Be careful while finding the exact solution of the equation as we have given that \[0\le \theta \le 2\pi \]. Also, be careful while solving the quadratic equation in \[\cos \theta \] as there is a chance that you might make a mistake while taking the common terms out. We can also find the exact solution by drawing the graph of \[\cos \theta =\dfrac{1}{2}\] for \[0\le \theta \le 2\pi \].