Question

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

Answer 1

Hint : The given trigonometric equation can be simplified using the trigonometric identities to get the equation that has only one trigonometric function. After, simplification uses the general solution of a trigonometric equation of the function and get all the solutions.
Formula used:
We have the general solution of the equation as $\sin x = \sin y$ is given by $x = n\pi + {( - 1)^n}y$ , where $n \in \mathbb{Z}$ .

Complete step-by-step answer :
The given equation is ${\left[ {\sin \left( {2x} \right) + \cos \left( {2x} \right)} \right]^2} = 1$ - - - - - - - - - - - - - - - - - (1)
Now, we will use the algebraic identity ${\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}$ to simplify the given equation as ${\sin ^2}\left( {2x} \right) + 2\sin \left( {2x} \right)\cos \left( {2x} \right) + {\cos ^2}\left( {2x} \right) = 1$ .
Rearranging the terms, we get, (1) as ${\sin ^2}\left( {2x} \right) + {\cos ^2}\left( {2x} \right) + 2\sin \left( {2x} \right)\cos \left( {2x} \right) = 1$ .
$\Rightarrow 1 + 2\sin \left( {2x} \right)\cos \left( {2x} \right) = 1$
[Using the identity ${\sin ^2}\theta + {\cos ^2}\theta = 1$ ]
$\Rightarrow 2\sin \left( {2x} \right)\cos \left( {2x} \right) = 1 - 1 = 0$
$\Rightarrow \sin \left( {4x} \right) = 0$
[Using the identity $\sin 2\theta = 2\sin \theta \cos \theta$ ]
$\Rightarrow \sin \left( {4x} \right) = \sin \left( 0 \right)$
[Since, $\sin \left( 0 \right) = 0$ ]
Now, we will use the general solution of the trigonometric equation $\sin x = \sin y$ is given by $x = n\pi + {( - 1)^n}y$ , where $n \in \mathbb{Z}$ .
$\Rightarrow 4x = n\pi + {( - 1)^n}\left( 0 \right),{\text{ }}n \in \mathbb{Z}$
$\Rightarrow x = \dfrac{{n\pi }}{4},{\text{ }}n \in \mathbb{Z}$ , is the general solution of (1).
$\Rightarrow x = 0,\dfrac{\pi }{4},\dfrac{\pi }{2},\dfrac{{3\pi }}{4},\pi ,\dfrac{{5\pi }}{4},\dfrac{{3\pi }}{2},\dfrac{{7\pi }}{4},2\pi ,...$ for $n = 0,1,2,3,4,5,6,7,8,...$
But to find the solutions in the given interval which is positive, we will look at the solutions generated by the non-negative integers.
So, the solutions for the given equation (1) in the interval $[0,2\pi ]$ are $0,\dfrac{\pi }{4},\dfrac{\pi }{2},\dfrac{{3\pi }}{4},\pi ,\dfrac{{5\pi }}{4},\dfrac{{3\pi }}{2},\dfrac{{7\pi }}{4}{\text{ and }}2\pi$ .
So, the correct answer is “ $0,\dfrac{\pi }{4},\dfrac{\pi }{2},\dfrac{{3\pi }}{4},\pi ,\dfrac{{5\pi }}{4},\dfrac{{3\pi }}{2},\dfrac{{7\pi }}{4}{\text{ and }}2\pi$ .”.

Note : Other important general solutions to solve the trigonometric equations are,
$\tan x = \tan y \Rightarrow x = n\pi + y,{\text{ where }}n \in \mathbb{Z}$
$\cos x = \cos y \Rightarrow x = 2n\pi \pm y,{\text{ where }}n \in \mathbb{Z}$
It is important to know all the trigonometric formulas and identities beforehand to simplify the trigonometric equation. So, that we can use the three known general solutions. It is very important to keep the interval in mind while solving the equation. Because that itself would make the steps simpler and can help in choosing the required solution.