Question

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

Answer 1

Hint: 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] $ .