Question

How do you find all the solutions of the following equation in the interval \[\left[ 0,2\pi \right]\] of \[2{{\cos }^{2}}x-\cos x=0\]?

Answer 1

Hint: This type of question is based on the concept of factorisation. First we have to simplify the given function by taking \[\cos x\] common. Then, we get two functions multiplied together to form the given equation, that is, \[\cos x\left( 2\cos x-1 \right)=0\] .thus, we get the factors of the equation. We then equate these factors to 0 and obtain the value of x using a trigonometric table which is the required answer.

Complete step-by-step answer:
According to the question, we are asked to find solutions of x for the function \[2{{\cos }^{2}}x-\cos x=0\] in the interval \[\left[ 0,2\pi \right]\].
We have been given the equation is \[2{{\cos }^{2}}x-\cos x=0\] --------(1)
We first have to simplify the given equation (1).
First, let us look for the common terms.
Here, we find \[\cos x\] common.
Therefore, on taking \[\cos x\] common from the equation (1), we get,
\[\cos x\left( 2\cos x-1 \right)=0\]
Here, \[\cos x\] and \[\left( 2\cos x-1 \right)\] are the factors of the given equation (1).
Since factors of the trigonometric equations are equal to 0, we get
\[\cos x=0\] and \[2\cos x-1=0\].
Let us now consider \[2\cos x-1=0\]. ---------(2)
Add 1 to both the sides of the equation (2).
We get,
\[\Rightarrow \cos \left( \dfrac{\left( 2n+1 \right)\pi }{2} \right)=0\dfrac{1}{2}\]
On further simplifications, we get,
\[2\cos x=1\]
Let us now divide the obtained expression by 2.
Therefore,
\[\dfrac{2\cos x}{2}=\dfrac{1}{2}\]
\[\Rightarrow \cos x=\dfrac{1}{2}\]
Here, we get cosx=0 and \[\cos x=\dfrac{1}{2}\].
We know that \[\cos \left( \dfrac{\pi }{2} \right)=0\], \[\cos \left( \dfrac{3\pi }{2} \right)=0\], \[\cos \left( \dfrac{5\pi }{2} \right)=0\] and so on.
From the above, we find a common pattern, that is, \[\cos \left( \dfrac{\left( 2n+1 \right)\pi }{2} \right)=0\].
Here n=0,1,2,…..
Similarly, we know that \[\cos \left( \dfrac{\pi }{6} \right)=\dfrac{1}{2}\], \[\cos \left( \dfrac{7\pi }{6} \right)=\dfrac{1}{2}\], \[\cos \left( \dfrac{13\pi }{6} \right)=\dfrac{1}{2}\] and so on.
From the above, we find a common pattern, that is, \[\cos \left( n\pi +\dfrac{\pi }{6} \right)=0\].
Here n=0,1,2,…..
But we have been asked to find the value of x in the interval \[\left[ 0,2\pi \right]\]
Therefore, the values of x for cosx=0 in the interval \[\left[ 0,2\pi \right]\] are \[\dfrac{\pi }{2},\dfrac{3\pi }{2}\].
And the values of x for \[\cos x=\dfrac{1}{2}\] in the interval \[\left[ 0,2\pi \right]\] are \[\dfrac{\pi }{6},\dfrac{7\pi }{6}\].
 Hence, the values of x for the given equation \[2{{\cos }^{2}}x-\cos x\left( 2\cos x-1 \right)=0\] in the interval \[\left[ 0,2\pi \right]\] are \[\dfrac{\pi }{6},\dfrac{\pi }{2},\dfrac{7\pi }{6}\] and \[\dfrac{3\pi }{2}\].

Note: We can also solve this type of questions by substituting cosx=u. Then we get a quadratic equation with the variable ‘u’, that is, \[2{{u}^{2}}-u=0\]. Solve this equation by taking u common and we get \[u\left( 2u-1 \right)=0\]. Now find the values of u, that is 0 and \[\dfrac{1}{2}\]. And convert u as cosx and follow the above mentioned steps to get the final answer.