Question

How do you find the exact solutions of the equation \[{\left( {\sin 2x + \cos 2x} \right)^2} = 1\] in the interval \[\left[ {0,2\pi } \right)\] ?

Answer 1

Hint: We are given with the expression and an interval also. Such that we are about to find the values of angle that satisfy the expression in the interval. So we will take the square identity to expand the expression. Then we will use the standard trigonometric identity to simplify the expression. After that we will be equating the expression and find the value of x.

Complete step by step solution:
Given that \[{\left( {\sin 2x + \cos 2x} \right)^2} = 1\]
We know that \[{\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}\]
Applying this on the above expression we get,
 \[{\left( {\sin 2x + \cos 2x} \right)^2} = {\left( {\sin 2x} \right)^2} + 2 \times \sin 2x \times \cos 2x + {\left( {\cos 2x} \right)^2}\]
We can rewrite this as,
 \[{\left( {\sin 2x + \cos 2x} \right)^2} = {\sin ^2}2x + 2 \times \sin 2x \times \cos 2x + {\cos ^2}2x\]
We know that \[{\sin ^2}x + {\cos ^2}x = 1\]
So we can write \[{\left( {\sin 2x + \cos 2x} \right)^2} = 1 + 2 \times \sin 2x \times \cos 2x\]
Equating this we get,
 \[1 = 1 + 2 \times \sin 2x \times \cos 2x\]
Taking 1 on other side we get,
 \[2 \times \sin 2x \times \cos 2x = 0\]
We know the identity that,
 \[2\sin \theta \cos \theta = \sin 2\theta \]
So we can write,
 \[\Rightarrow 2 \times \sin 2x \times \cos 2x = \sin 4x\]
Equating this to zero we get,
 \[\Rightarrow \sin 4x = 0\]
We know that \[\sin n\pi = 0\]
So we can say that \[x = \dfrac{{n\pi }}{4}\]
From the interval we get \[x = \dfrac{\pi }{4},\pi \]
These are the values of the angle for the given expression.
So, the correct answer is “ \[x = \dfrac{\pi }{4},\pi \] ”.

Note: Here note that the interval is already given so we have a limit to write the value of the angle. The meaning or notations of the limit is it is open if having a parenthesis and it is closed if it is having a square bracket. There are limits given in the case above. So we took particular values. We checked for various values of \[\dfrac{{n\pi }}{4}\] . From them we came to get the values noted above because the other values are not correct for the value of the function.