Question

Solve the trigonometric equation $ 2{\cos ^2}\left( x \right) = 1 $ in the interval $ [0,2\pi ] $

Answer 1

Hint: The given question involves solving a trigonometric equation and finding the value of angle x that satisfy the given equation and lie in the range of $ [0,2\pi ] $ . There can be various methods to solve a specific trigonometric equation. For solving such questions, we need to have knowledge of basic trigonometric formulae and identities.

Complete step-by-step answer:
In the given problem, we have to solve the trigonometric equation $ 2{\cos ^2}\left( x \right) = 1 $ and find the values of x that satisfy the given equation and lie in the range of $ [0,2\pi ] $ .
So, In order to solve the given trigonometric equation $ 2{\cos ^2}\left( x \right) = 1 $ , we should first take all the terms to the left side of the equation.
Transposing all the terms to left side of the equation, we get,
 $ = 2{\cos ^2}\left( x \right) - 1 = 0 $
Now, we know the double angle formula for cosine,
$ 2{\cos ^2}\left( x \right) - 1 = \cos \left( {2x} \right) $ . Hence, substituting $ \left[ {2{{\cos }^2}\left( x \right) - 1} \right] $ as $ \cos \left( {2x} \right) $ , we get,
 $ = \cos \left( {2x} \right) = 0 $
The above equation represents a simple form of the trigonometric equation. We know that cosine is equal to zero at odd multiples of $ \left( {\dfrac{\pi }{2}} \right) $ .
So, $ 2x = \left( {2n + 1} \right)\left( {\dfrac{\pi }{2}} \right) $
 $ = x = \left( {2n + 1} \right)\left( {\dfrac{\pi }{4}} \right) $
Now, we have found all the values of x that satisfy the given trigonometric equation. Now, we just have to select the values of that lie in the interval $ [0,2\pi ] $ .
So, for $ n = 0 $ in $ x = \left( {2n + 1} \right)\left( {\dfrac{\pi }{4}} \right) $ , we get $ x = \dfrac{\pi }{4} $ .
For $ n = 1 $ in $ x = \left( {2n + 1} \right)\left( {\dfrac{\pi }{4}} \right) $ , we get $ x = \dfrac{{3\pi }}{4} $ .
For $ n = 2 $ in $ x = \left( {2n + 1} \right)\left( {\dfrac{\pi }{4}} \right) $ , we get $ x = \dfrac{{5\pi }}{4} $ .
For $ n = 3 $ in $ x = \left( {2n + 1} \right)\left( {\dfrac{\pi }{4}} \right) $ , we get $ x = \dfrac{{7\pi }}{4} $ .
Hence, the values of x that satisfy the given trigonometric equation $ 2{\cos ^2}\left( x \right) = 1 $ and lie between the interval $ [0,2\pi ] $ are: $ x = \dfrac{\pi }{4} $ , $ \dfrac{{3\pi }}{4} $ , $ \dfrac{{5\pi }}{4} $ and $ \dfrac{{7\pi }}{4} $ .
So, the correct answer is “ $ x = \dfrac{\pi }{4} $ , $ \dfrac{{3\pi }}{4} $ , $ \dfrac{{5\pi }}{4} $ and $ \dfrac{{7\pi }}{4} $ ”.

Note: The given trigonometric equation can also be solved by first finding the value of $ {\cos ^2}\left( x \right) $ in $ 2{\cos ^2}\left( x \right) = 1 $ as ${\cos ^2}\left( x \right) = \dfrac{1}{2} $ and then finding the value of $ \cos \left( x \right) $ as $ \left( { \pm \dfrac{1}{{\sqrt 2 }}} \right) $ . Then, we solve the two equations $ \cos \left( x \right) = \dfrac{1}{{\sqrt 2 }} $ and $ \cos \left( x \right) = - \dfrac{1}{{\sqrt 2 }} $ and find the values of x that satisfy either of the equations and lie between the interval $ [0,2\pi ] $ .