Question

How do you solve $2{\cos ^3}x + {\cos ^2}x = 0$ in the interval $[0,2\pi ]$ ?

Answer 1

Hint : Before solving the problem we need to observe that the given interval is a closed interval and we can include 0 and $2\pi$ . Now we have $2{\cos ^3}x + {\cos ^2}x = 0$ . In this we can take ${\cos ^2}x$ as common and grouping we have two factors. Solving these factors we will get the required answer.

Complete step-by-step answer :
Given,
$2{\cos ^3}x + {\cos ^2}x = 0$
Now divide the whole equation by 2
${\cos ^3}x + \dfrac{1}{2}{\cos ^2}x = 0$ .
Now take ${\cos ^2}x$ common we have,
${\cos ^2}x\left( {\cos x + \dfrac{1}{2}} \right) = 0$
Now using zero product principle we have,
${\cos ^2}x = 0$ and $\cos x + \dfrac{1}{2} = 0$
$\cos x = 0$ and $\cos x = - \dfrac{1}{2}$ .
Since we have two factors, let’s take the first factor.
That is $\cos x = 0$ .
$\cos x = \cos \dfrac{\pi }{2}$
Since the period of cosine is $2\pi$ and the values will repeat for every period. That is
$x = 2k\pi \pm \dfrac{\pi }{2}$ .
This is the general solution of $\cos x = 0$
Put $k = 0$ in the above general solution. then
$x = \pm \dfrac{\pi }{2}$ .
Put $k = 1$ in above general solution we have,
$x = 2\pi \pm \dfrac{\pi }{2}$
That is $x = 2\pi + \dfrac{\pi }{2} = \dfrac{{5\pi }}{2}$ and $x = 2\pi - \dfrac{\pi }{2} = \dfrac{{3\pi }}{2}$ .
Since $x \in [0,2\pi ]$ . The solution of $\cos x = 0$ for $x \in [0,2\pi ]$ is $\dfrac{\pi }{2}$ and $\dfrac{{3\pi }}{2}$ .
(because remaining values are not in the closed intervals)
Now take the second factor.
That is $\cos x = - \dfrac{1}{2}$
But we know that at $x = \dfrac{{2\pi }}{3}$ and $x = \dfrac{{4\pi }}{3}$ the cosine value is $- \dfrac{1}{2}$ both are belongs in closed interval.
$\cos \left( {\dfrac{{2\pi }}{3}} \right) = \cos \left( {\pi - \dfrac{\pi }{3}} \right)$
We know the supplementary angle. $\cos (\pi - \theta ) = - \cos (\theta )$ , negative sign is because cosine lies in the second quadrant and cosine negative in the second quadrant.
$= - \cos \left( {\dfrac{\pi }{3}} \right)$
$= - \dfrac{1}{2}$
Similarly
$\cos \left( {\dfrac{{4\pi }}{3}} \right) = \cos \left( {\pi + \dfrac{\pi }{3}} \right)$
We know the supplementary angle. $\cos (\pi + \theta ) = - \cos (\theta )$ , negative sign is because cosine lies in the third quadrant and cosine negative in the third quadrant.
$= - \cos \left( {\dfrac{\pi }{3}} \right)$
$= - \dfrac{1}{2}$
Thus the solution of $2{\cos ^3}x + {\cos ^2}x = 0$ is $\dfrac{\pi }{2}$ , $\dfrac{{3\pi }}{2}$ , $\dfrac{{2\pi }}{3}$ and $\dfrac{{4\pi }}{3}$
So, the correct answer is “ $\dfrac{\pi }{2}$ , $\dfrac{{3\pi }}{2}$ , $\dfrac{{2\pi }}{3}$ and $\dfrac{{4\pi }}{3}$ ”.

Note : Since the period of cosine is $2\pi$ and even though if we take the solution of second factor as $x = \dfrac{{2\pi }}{3} + 2k\pi$ and $x = \dfrac{{4\pi }}{3} + 2k\pi$ for values of ‘k’ greater than 1, the obtained value does not belongs in the closed interval $[0,2\pi ]$ . Remember the signs changes of all the six trigonometric in four quadrants.