Question

How do you solve $\cos 2x = \cos x$ ?

Answer 1

In order to determine the solution of the above trigonometric equation, Rewrite the equation using trigonometric identity $\cos 2x = 2{\cos ^2}x - 1$ . You will obtain a quadratic equation, compare it with the standard equation to get the value of variables $a,b,c$ . Since $a + b + c = 0$ , obtain the solution by putting $\cos x = 0$ and $\cos x = \dfrac{c}{a} = - \dfrac{1}{2}$ .

Complete step by step answer:
We are given a trigonometric equation $\cos 2x = \cos x$ and we have to find its solution
Since in the question we have not given any constraint for the value of $x$ , so we will be finding the solution from $0$ to $2\pi$
We are going to rewrite our using the identity of trigonometry $\cos 2x = 2{\cos ^2}x - 1$
$2{\cos ^2}x - 1 = \cos x \\\ 2{\cos ^2}x - \cos x - 1 = 0 \\\$
Let $X = \cos x$
As we can see we obtained a quadratic equation, lets compare the above equation with the standard quadratic equation $a{X^2} + bX + c = 0$ , we get
$a = 2 \\\ b = - 1 \\\ c = - 1 \\\$
Since $a + b + c = 2 - 1 - 1 = 0$ , we can say that there are two real roots as
a. $\cos x = 0$ , $\Rightarrow x = {\cos ^{ - 1}}\left( 0 \right)$ $\Rightarrow x = 0\,or\,2\pi$
b. $\cos x = \dfrac{c}{a} = - \dfrac{1}{2}$ , $\Rightarrow x = {\cos ^{ - 1}}\left( { - \dfrac{1}{2}} \right) \Rightarrow x = \dfrac{{2\pi }}{3}\,or\,\dfrac{{4\pi }}{3}$

Therefore, the solution of given trigonometric equation is $x = 0,\dfrac{{2\pi }}{3},\,\,\dfrac{{4\pi }}{3},2\pi$ between $0$ to $2\pi$ .

Additional Information:
1. Trigonometry is one of the significant branches throughout the entire existence of mathematics and this idea is given by a Greek mathematician Hipparchus.
2. Even Function – A function $f(x)$ is said to be an even function ,if $f( - x) = f(x)$ for all x in its domain.
Odd Function – A function $f(x)$ is said to be an even function ,if $f( - x) = - f(x)$ for all x in its domain.
We know that $\sin ( - \theta ) = - \sin \theta .\cos ( - \theta ) = \cos \theta \,and\,\tan ( - \theta ) = - \tan \theta$
Therefore, $\sin \theta$ and $\tan \theta$ their reciprocals, $\cos ec\theta$ and $\cot \theta$ are odd functions whereas $\cos \theta$ and its reciprocal $\sec \theta$ are even functions.
3. Periodic Function= A function $f(x)$ is said to be a periodic function if there exists a real number T > 0 such that $f(x + T) = f(x)$ for all x.

Note: 1.One must be careful while taking values from the trigonometric table and cross-check at least once to avoid any error in the answer.
2. Period of cosine function is $2\pi$ .
3. The domain of cosine function is in the interval $\left[ {0,\pi } \right]$ and the range is in the interval $\left[ { - 1,1} \right]$ .