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 $[0,2\pi ]$ .