Question

# Solve the following equation:$2{{\sin }^{2}}\theta =3\cos \theta$ in the interval $0\le \theta \le 2\pi$.

Hint: Use the formula ${{\sin }^{2}}x=1-{{\cos }^{2}}x$ in the given equation and convert the equation into quadratic equation in the form of $\cos \theta$. Find the solution of the quadratic equation and use the concept “If $\cos x=\cos y$ then, $x=2n\pi \pm y$” so that you can get the answer in the interval of $0\le \theta \le 2\pi$.

To solve the above equation we have to write it down therefore,
$2{{\sin }^{2}}\theta =3\cos \theta$
As we can see the ${{\sin }^{2}}\theta$ can easily be converted to ${{\cos }^{2}}\theta$ by using the formula given below so that we will find easier to solve the equation in $\cos \theta$.
Formula:
${{\sin }^{2}}x=1-{{\cos }^{2}}x$
If we use the above formula in the given equation w will get,
$\therefore 2\left( 1-{{\cos }^{2}}\theta \right)=3\cos \theta$
If we multiply by ‘2’ inside the bracket we will get,
$\therefore 2-2{{\cos }^{2}}\theta =3\cos \theta$
If we shift $3\cos \theta$ on the left hand side of the equation we will get,
$\therefore 2-2{{\cos }^{2}}\theta -3\cos \theta =0$
Now we will multiply the above equation by ‘-1’ to get,
$\therefore \left( -1 \right)\times \left( 2-2{{\cos }^{2}}\theta -3\cos \theta \right)=0$
$\therefore -2+2{{\cos }^{2}}\theta +3\cos \theta =0$
By rearranging the above equation we will get,
$\therefore 2{{\cos }^{2}}\theta +3\cos \theta -2=0$
Now we can write $3\cos \theta$as $\left( -4\cos \theta +\cos \theta \right)$ therefore we will get,
$\therefore 2{{\cos }^{2}}\theta +\left( -4\cos \theta +\cos \theta \right)-2=0$
$\therefore 2{{\cos }^{2}}\theta -4\cos \theta +\cos \theta -2=0$
Taking $2\cos \theta$ common from above equation we will get,
$\therefore 2\cos \theta \left( \cos \theta -2 \right)+\cos \theta -2=0$
If we rearrange the above equation we will get,
$\therefore 2\cos \theta \left( \cos \theta -2 \right)+\left( \cos \theta -2 \right)=0$
Also by taking $\left( \cos \theta -2 \right)$ from the above equation we will get,
$\therefore \left( \cos \theta -2 \right)\left( 2\cos \theta +1 \right)=0$
As they are in multiplication therefore we can write,
$\left( \cos \theta -2 \right)=0$ ………………………….. (1)
OR
$\left( 2\cos \theta +1 \right)=0$………………………….. (2)
We will write the equation (1) first, therefore,
$\left( \cos \theta -2 \right)=0$
$\therefore \cos \theta =2$
But as we know that the range of $\cos \theta$ is [-1, 1] Therefore the value of $\cos \theta$ equal to 2 is not possible, therefore we can write,
$\therefore \cos \theta =2$ Not possible.
Hence rejected.
Now we will solve the equation (2), therefore,
$\left( 2\cos \theta +1 \right)=0$
$\therefore 2\cos \theta =-1$
$\therefore \cos \theta =\dfrac{-1}{2}$
$\therefore \cos \theta =-\dfrac{1}{2}$
As we know that the value of $\cos \dfrac{\pi }{3}=\dfrac{1}{2}$ and if we put this value in above equation we will get,
$\therefore \cos \theta =-\cos \dfrac{\pi }{3}$
Now to proceed further in the solution we should know the formula given below,
Formula:
$-\cos x=\cos \left( \pi -x \right)$
If we use the formula in the equation we will get,
$\therefore \cos \theta =\cos \left( \pi -\dfrac{\pi }{3} \right)$
$\therefore \cos \theta =\cos \left( \dfrac{3\pi -\pi }{3} \right)$
$\therefore \cos \theta =\cos \left( \dfrac{2\pi }{3} \right)$
To proceed further in the solution we should know the formula given below,
Formula:
If $\cos x=\cos y$ then, $x=2n\pi \pm y$
If we use the above formula in the equation we will get,
As, $\cos \theta =\cos \left( \dfrac{2\pi }{3} \right)$ therefore, $\theta =2n\pi \pm \dfrac{2\pi }{3}$
As we have to find the solution between $0\le \theta \le 2\pi$ therefore we will put n = 1 in the solution, therefore we will get,
$\therefore \theta =2\left( 1 \right)\pi \pm \dfrac{2\pi }{3}$
This can also be written as,
$\therefore \theta =2\pi +\dfrac{2\pi }{3}$ And $\therefore \theta =2\pi -\dfrac{2\pi }{3}$
If we do the required calculations we will get,
$\therefore \theta =\dfrac{6\pi +2\pi }{3}$ And $\therefore \theta =\dfrac{6\pi -2\pi }{3}$
$\therefore \theta =\dfrac{8\pi }{3}$ And $\therefore \theta =\dfrac{4\pi }{3}$
As we know that $\dfrac{8\pi }{3}>2\pi$ therefore the answer is,
$\therefore \theta =\dfrac{4\pi }{3}$
$\therefore \theta =\dfrac{2\pi }{3},\dfrac{4\pi }{3}$
Therefore the solution of $2{{\sin }^{2}}\theta =3\cos \theta$ in the interval $0\le \theta \le 2\pi$ is $\dfrac{2\pi }{3}$ and $\dfrac{4\pi }{3}$.
Note: From the step $\cos \theta =-\dfrac{1}{2}$ you can also solve further by using the formula of inverse trigonometric functions i.e. ${{\cos }^{-1}}\left( -x \right)=\pi ={{\cos }^{-1}}x$ and you can further manage the solution in the given interval.