Question

Solve this \[2{{\sin }^{2}}\left( x \right)-{{\sin }^{2}}\left( 2x \right)={{\cos }^{2}}\left( 2x \right)\]. It should be: \[x=\dfrac{\pi }{4}+k\pi \] and \[x=\dfrac{3\pi }{4}+k\pi \]

Answer 1

Hint: In order to find the general solution of the given trigonometric equation that is \[2{{\sin }^{2}}\left( x \right)-{{\sin }^{2}}\left( 2x \right)={{\cos }^{2}}\left( 2x \right)\], simplify this equation with the help of trigonometric property of sine and cosine that is \[{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1\]. After that, take the square root of both the sides of the equation and you’ll get two values of \[\sin \left( x \right)\]. Now consider two cases for these values and find the value of \[x\] in each case with the help of the trigonometric property of sine and sine inverse that is \[{{\sin }^{-1}}\left( \sin \left( x \right) \right)=x\]

Complete step by step solution:
Given equation is as follows:
\[2{{\sin }^{2}}\left( x \right)-{{\sin }^{2}}\left( 2x \right)={{\cos }^{2}}\left( 2x \right)\]
Add \[{{\sin }^{2}}\left( 2x \right)\] both the sides of the above equation we get:
\[\Rightarrow 2{{\sin }^{2}}\left( x \right)={{\sin }^{2}}\left( 2x \right)+{{\cos }^{2}}\left( 2x \right)\]
Apply the trigonometric property of sine and cosine that is \[{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1\]in the above equation we get:
\[\Rightarrow 2{{\sin }^{2}}\left( x \right)=1\]
Now after dividing 2 both the sides from the above equation, we get:
\[\Rightarrow {{\sin }^{2}}\left( x \right)=\dfrac{1}{2}\]
Taking square root on both the sides we get:
\[\Rightarrow \sin \left( x \right)=\pm \dfrac{1}{\sqrt{2}}\]
This implies there are two values of \[\sin \left( x \right)\] for which \[2{{\sin }^{2}}\left( x \right)-{{\sin }^{2}}\left( 2x \right)={{\cos }^{2}}\left( 2x \right)\] this equation is satisfied. So, consider two cases to find the value of \[x\] in each case.
Case 1: \[\sin \left( x \right)=\dfrac{1}{\sqrt{2}}\] that is Positive value;
Take inverse sine inverse both the sides we get:
\[\Rightarrow {{\sin }^{-1}}\left( \sin \left( x \right) \right)={{\sin }^{-1}}\left( \dfrac{1}{\sqrt{2}} \right)\]
Apply the trigonometric property of sine and sine inverse that is \[{{\sin }^{-1}}\left( \sin \left( x \right) \right)=x\] in the above equation we get:
\[\Rightarrow x={{\sin }^{-1}}\left( \dfrac{1}{\sqrt{2}} \right)\]
We know that value of \[{{\sin }^{-1}}\left( \dfrac{1}{\sqrt{2}} \right)=\dfrac{\pi }{4}\]. Therefore, we can write
\[\Rightarrow x=\dfrac{\pi }{4}+k\pi \] where \[k\] is even.
Case 2: \[\sin \left( x \right)=-\dfrac{1}{\sqrt{2}}\] that is negative value;
Take inverse sine inverse both the sides, we get:
\[\Rightarrow {{\sin }^{-1}}\left( \sin \left( x \right) \right)={{\sin }^{-1}}\left( -\dfrac{1}{\sqrt{2}} \right)\]
Apply the trigonometric property of sine and sine inverse that is \[{{\sin }^{-1}}\left( \sin \left( x \right) \right)=x\] in the above equation we get:
\[\Rightarrow x={{\sin }^{-1}}\left( -\dfrac{1}{\sqrt{2}} \right)\]
We know that value of \[{{\sin }^{-1}}\left( \dfrac{1}{\sqrt{2}} \right)=\dfrac{3\pi }{4}\]. Therefore, we can write
\[\Rightarrow x=\dfrac{3\pi }{4}+k\pi \] where \[k\] is odd.
Therefore, General solution of the equation \[2{{\sin }^{2}}\left( x \right)-{{\sin }^{2}}\left( 2x \right)={{\cos }^{2}}\left( 2x \right)\] are \[x=\dfrac{\pi }{4}+k\pi \] when \[k\] is even and \[x=\dfrac{3\pi }{4}+k\pi \] when \[k\] is odd.

Note:
Students can go wrong while taking the square root of \[{{\sin }^{2}}\left( x \right)=\dfrac{1}{2}\] and take only positive value as the answer and ignoring the negative value that is instead of writing this \[\sin \left( x \right)=\pm \dfrac{1}{\sqrt{2}}\]as the square root of \[{{\sin }^{2}}\left( x \right)=\dfrac{1}{2}\], some students write this \[\sin \left( x \right)=\dfrac{1}{\sqrt{2}}\]only which is partially correct but leads to the incomplete answer. Key point is to remember whenever taking square roots consider both positive and negative values, especially when you are finding a general solution to some question.