Question

Solve the trigonometric expression $ \cos 2x = 1 $ and find the value of $ x $ .

Answer 1

Hint: We are given a trigonometric equation and we have to find the value of $ x $ in the range of cosine function so we use the $ \cos 2x $ formula to simplify and solve the trigonometric expression or we can equate the equations and by comparison find the answer.

Complete step-by-step answer:
Firstly we write down the trigonometric equation given in the question
 $ \cos 2x = 1 $
We have two methods to solve this equation and find the value of $ x $ . In the first method we use the standard trigonometric identity of $ \cos 2x $ and find the value of $ x $ serving our purpose. In the second method we equate the equation by making both sides into cosine function and then equating the angles to get the answer.
Using first method we substitute the given formula of
 $ \cos 2x = 2{\cos ^2}x - 1 $
Into the given equation and we have
 \[
  \cos 2x = 2{\cos ^2}x - 1 = 1 \\
   \Rightarrow 2{\cos ^2}x = 1 + 1 = 2 \\
   \Rightarrow {2}{\cos ^2}x = {2} \;
 \]
We have got the value of cosine function as
 $
  {\cos ^2}x = 1 \\
   \Rightarrow \cos x = \pm 1 \;
  $
So we have got the value of the cosine function i.e. $ \pm 1 $ . As we know that
 $
  \cos 0^\circ = 1 \\
  \cos 2\pi = 1 \\
  \cos \pi = - 1 \;
  $
This means all three angles $ 0^\circ ,2\pi \,{\text{and}}\,\pi $ satisfies the equation and hence
 $ x = 0,\pi ,2\pi $
In method second we convert the RHS part of the equation into a trigonometric expression i.e.
 $ 1 = \cos 0,\cos 2\pi $
So when we substitute these values into the given equation we have
 $
  \cos 2x = \cos 0 \\
   \Rightarrow x = 0 \\
  \cos 2x = \cos 2\pi \\
   \Rightarrow x = \pi ;
  $
We have got the same result using this method too.
So, the correct answer is “$x = 0,\pi ,2\pi $ ”.

Note: All multiples of $ 0\,{\text{and}}\,\pi $ will be true for the value of $ x $ in this question i.e. $ 0,\pi ,2\pi ,3\pi ,.... $ because in the end the angle will be multiplied by 2 making the angle lying on the same line ultimately where the value of cosine is always $ 1 $