Question

How do you solve $\sin 4x - 2\sin 2x = 0$ in the interval $\left[ {0,360} \right]$?

Answer 1

Hint: This problem deals with solving the given equation with trigonometric identities and compound sum angles of trigonometric functions. A compound angle formula or addition formula is a trigonometric identity which expresses a trigonometric function of $\left( {A + B} \right)$ or $\left( {A - B} \right)$in terms of trigonometric functions of $A$ and $B$. The used formula here is:
$ \Rightarrow \sin \left( {A + B} \right) = \sin A\cos B + \cos A\sin B$

Complete step-by-step solution:
Given an equation of trigonometric expression functions.
The given equation is $\sin 4x - 2\sin 2x = 0$, consider this as given below:
$ \Rightarrow \sin 4x - 2\sin 2x = 0$
We know that by using one of the trigonometric identity, as given below:
Consider only $\sin 4x$, applying the sine trigonometric compound angle to it, as given below:
$ \Rightarrow \sin 4x = \sin \left( {2x + 2x} \right)$
$ \Rightarrow \sin 4x = \sin 2x\cos 2x + \cos 2x\sin 2x$
On further simplifying the above expression, as given below:
$ \Rightarrow \sin 4x = 2\sin 2x\cos 2x$
Substitute this expression in the given equation of $\sin 4x - 2\sin 2x = 0$, as given below:
$ \Rightarrow 2\sin 2x\cos 2x - 2\sin 2x = 0$
From the above expression taking the term $2\sin 2x$ common, as given below:
$ \Rightarrow 2\sin 2x\left( {\cos 2x - 1} \right) = 0$
Now $2\sin 2x$ should be zero, or $\cos 2x - 1$ should be equal to zero, so considering both the cases, as shown below:
First consider $\sin 2x$ is equated to zero, as given below:
$ \Rightarrow \sin 2x = 0$
$ \Rightarrow 2x = 0,180,360,540,720$
To get the value of $x$, dividing the above expression by 2, as shown below:
$ \Rightarrow x = 0,\dfrac{{180}}{2},\dfrac{{360}}{2},\dfrac{{540}}{2},\dfrac{{720}}{2}$
$\therefore x = 0,90,180,270,360$
Now consider $\cos 2x - 1$ is equated to zero, as given below:
$ \Rightarrow \cos 2x - 1 = 0$
$ \Rightarrow \cos 2x = 1$
$ \Rightarrow 2x = 0,360,720$
To get the value of $x$, dividing the above expression by 2, as shown below:
$ \Rightarrow x = 0,\dfrac{{360}}{2},\dfrac{{720}}{2}$
$\therefore x = 0,180,360$
Now combining the values of $x$, from both the cases, as shown below:
$ \Rightarrow x = \left\{ {0,90,180,270,360} \right\} \cup \left\{ {0,180,360} \right\}$
$\therefore x = \left\{ {0,90,180,270,360} \right\}$

The values of x in the interval are $\left\{ {0,90,180,270,360} \right\}$.

Note: Please note that while solving the values of $x$, in both the cases where $\sin 2x = 0$ and $\cos 2x = 1$, here the solutions for $x$ can be infinite number of solutions in both the cases, but here we are restricted with the given interval $\left[ {0,360} \right]$, hence the solutions for $x$, are limited to the given interval.