Question

Solve the equation, \[2{{\cos }^{2}}\theta -5\cos \theta +2=0\].

Answer 1

Hint: At first try to factorize the given quadratic equation and find values of \[\cos \theta \] for which the equation satisfies. Then find the general solution of cases which are valid or possible.

Complete step-by-step answer:

In the question we are given an equation which is \[2{{\cos }^{2}}\theta -5\cos \theta +2=0\]

and we have to find a general solution or values of \[\theta \].

So, we are given that,

\[2{{\cos }^{2}}\theta -5\cos \theta +2=0\]

Now, we can rewrite the equation as,

\[2{{\cos }^{2}}\theta -4\cos \theta -\cos \theta +2=0\]

We will factorize now the given equation and write it as,

\[2\cos \theta \left( \cos \theta -2 \right)-1\left( \cos \theta -2 \right)=0\]

Or, \[\left( 2\cos \theta -1 \right)\left( \cos \theta -2 \right)=0\]

So, satisfying the given equation the value of \[\cos \theta \] should be equal to either \[\dfrac{1}{2}\] or 2.

So, let’s first take the case of \[\cos \theta =\dfrac{1}{2}\].

We know that, \[\cos \dfrac{\pi }{3}=\dfrac{1}{2}\]. So, we can write \[\dfrac{1}{2}\] as \[\cos \dfrac{\pi }{3}\] in the equation.

So, we can write it as,

\[\cos \theta =\cos \dfrac{\pi }{3}\]

So, \[\theta =2n\pi \pm \dfrac{\pi }{3}\], where n is any integer using formula that if \[\cos \theta =\cos \alpha \], so, \[\theta =2n\pi \pm \alpha \], where, \[\alpha \in \left[ \dfrac{-\pi }{2},\dfrac{\pi }{2} \right]\].

Now, for the other case where, \[\cos \theta =2\].

This case is not possible as the value of \[\cos \theta \] lies between -1 and 1 so it will not attain value 2.

Hence the value of \[\theta \] is \[2n\pi \pm \dfrac{\pi }{3}\].

Note: Instead of doing like this one can also solve the equation by taking \[\cos \theta =t\] and then use the quadratic equation formula to find roots and hence substituting t as \[\cos \theta \] and find general solutions.