Question

How do you find the exact solutions of $\cos 2x-\cos x=0$ in the interval $\left[0,2\pi \right)$?

Answer 1

Hint:
Here we can turn the above given equation into the quadratic equation in $\cos x$ and then we can find the different values of the $\cos x$ and then according to its value we can find the value of $x$ from the graph of the $\cos x$

Complete step by step solution:
Here we need to find the value of $x$ but in the interval which is given as $\left[0,2\pi \right)$
In this interval we must know that this open bracket means that $2\pi$ is not included in the interval and towards the left we have the closed bracket which means $0$ is included in the interval. Hence we cannot take $2\pi$ as our answer.
So here we are given the equation as:
$\cos 2x-\cos x=0$$---(1)$
We know that $\cos 2x={\cos }^{2}x-{\sin }^{2}x$$-----(2)$
Also we know that
$$\begin{gathered} {\sin }^{2}x+{\cos }^{2}x=1 \\ {\sin }^{2}x=1-{\cos }^{2}x \end{gathered}$$
Now we can substitute this value in the equation (2) and get:
$$\begin{gathered} \cos 2x={\cos }^{2}x-(1-{\cos }^{2}x) \\ \cos 2x=2{\cos }^{2}x-1-----(3) \end{gathered}$$
Now substituting this value we get in equation (3) in the equation (1) we will get:
$$\begin{gathered} 2{\cos }^{2}x-1-\cos x=0 \\ 2{\cos }^{2}x-\cos x-1=0 \end{gathered}$$
Now we get the quadratic equation in $\cos x$
Now we can write in above equation that $\left(-\cos x\right)=\left(-2\cos x+\cos x\right)$
We will get:
$2{\cos }^{2}x-2\cos x+\cos x-1=0$
Simplifying it further we will get:
$2\cos x(\cos x-1)+(\cos x-1)=0$
$\left(2\cos x+1\right)\left(\cos x-1\right)=0$
So we can say either $\cos x=-{\displaystyle \frac{1}{2}}\text{ or }\cos x=1$
Now we can plot the graph of $\cos x$ which is as:

Now we know that from the graph we can see:
 $\cos x=-{\displaystyle \frac{1}{2}}\text{ or }\cos x=1$
$$\begin{gathered} \cos x=1 \\ x=0 \end{gathered}$$
$\cos x=-{\displaystyle \frac{1}{2}}$
From the graph we can notice that:
For $\cos x=-{\displaystyle \frac{1}{2}}$
$x={\displaystyle \frac{2\pi }{3}},{\displaystyle \frac{4\pi }{3}}$

Hence we get the values as $x=0,{\displaystyle \frac{2\pi }{3}},{\displaystyle \frac{4\pi }{3}}$

Note:
If we do not know the graph we can use the properties of cosine function which says that:
$\cos {\displaystyle \frac{2\pi }{3}}=\cos \left({\displaystyle \frac{\pi }{2}}+{\displaystyle \frac{\pi }{6}}\right)$
Now we know that $\cos \left({\displaystyle \frac{\pi }{2}}+x\right)=-\sin x$
So we get $\cos {\displaystyle \frac{2\pi }{3}}=\cos \left({\displaystyle \frac{\pi }{2}}+{\displaystyle \frac{\pi }{6}}\right)=-\sin {\displaystyle \frac{\pi }{6}}=-{\displaystyle \frac{1}{2}}$
Hence we must know the properties of all the trigonometric functions in order to solve such problems.