Question

How do you find all the solutions in the interval $\left[ {0,2\pi } \right]$ :$2{\cos ^2}\left( {2x} \right) - 1 = 0$ ?

Answer 1

Hint:
For this type of questions first isolate the given angle by following the reverse order of the operations. And solve for x and the values lie between given intervals.

Complete step by step solution:
The objective of the problem is to find all the solutions in the interval $\left[ {0,2\pi } \right]$ : $2{\cos ^2}\left( {2x} \right) - 1 = 0$.
To solve this we first isolate the angle given angle. The given angle is 2x. so we are isolating the angle 2x.
Given $2{\cos ^2}\left( {2x} \right) - 1 = 0$
To solve this follow the below procedure.
Add one on both sides in equation $2{\cos ^2}\left( {2x} \right) - 1 = 0$. We get
$
  2{\cos ^2}\left( {2x} \right) - 1 + 1 = 0 + 1 \\
   \Rightarrow 2{\cos ^2}\left( {2x} \right) = 1 \\
 $
Divide the above obtained equation with two on both sides we get
$
   \Rightarrow \dfrac{{2{{\cos }^2}\left( {2x} \right)}}{2} = \dfrac{1}{2} \\
   \Rightarrow {\cos ^2}\left( {2x} \right) = \dfrac{1}{2} \\
 $
Now taking square root on both sides of the above equation we get
$
   \Rightarrow \sqrt {{{\cos }^2}\left( {2x} \right)} = \sqrt {\dfrac{1}{2}} \\
   \Rightarrow \cos \left( {2x} \right) = \pm \dfrac{1}{{\sqrt 2 }} \\
 $
Therefore , $\cos \left( {2x} \right) = \dfrac{1}{{\sqrt 2 }}\,\,,\,\,\cos \left( {2x} \right) = - \dfrac{1}{{\sqrt 2 }}$
Now taking the inverse cosine functions to find the angle values . we get
${\cos ^{ - 1}}\left( {\cos \left( {2x} \right)} \right) = {\cos ^{ - 1}}\left( {\dfrac{1}{{\sqrt 2 }}} \right)\,\,,\,\,{\cos ^{ - 1}}\left( {\cos \left( {2x} \right)} \right) = {\cos ^{ - 1}}\left( { - \dfrac{1}{{\sqrt 2 }}} \right)$
Solve the above inverse functions we get
$2x = \dfrac{\pi }{4},\dfrac{{3\pi }}{4},\dfrac{{5\pi }}{4},\dfrac{{7\pi }}{4}...$
The value of cosine functions is $\dfrac{1}{{\sqrt 2 }}$ at the odd values of pi by four.
And negative of cosine theta is cosine theta.
But we need find the value of x. for this divide the obtained equation by two we get
 $x = \dfrac{\pi }{8},\dfrac{{3\pi }}{8},\dfrac{{5\pi }}{8},\dfrac{{7\pi }}{8}...$

Therefore the value of x lies between the given interval $\left[ {0,2\pi } \right]$.

Note:
Remember for this type of question we should follow the reverse order of the operations to isolate the values of given angle. While solving the quadratic form that is square form in the result we get the plus or minus values. The period of sine function is two pi and the period of cosine function is the same as the sine function. The period of tan function is pi.