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.