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

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.