Question

Solve the following trigonometric equation : $\sin 2x-\sin 4x+\sin 6x=0$.

Answer 1

Hint: Pair $\sin 2x$ with $\sin 6x$ and then use product to sum formula which is $\sin a+\sin b=2\sin \left( \dfrac{a+b}{2} \right)\cos \left( \dfrac{a-b}{2} \right)$, then take common terms out and equate the product of all the terms equal to 0 to find the general solution of the given trigonometric equation.

Complete step-by-step answer:
Trigonometric equations are equations involving trigonometric functions. A trigonometric equation that holds true for any angle is called a trigonometric identity. There are other equations that are true for certain angles. They are generally known as conditional equations. Solution of trigonometric equations requires concepts of different trigonometric identities like conversion from product to sum of trigonometric functions using product to sum rule and similarly conversion from sum to product using sum to product rule.

Now, when we talk about the solutions of trigonometric functions, there are two types of solutions: general solution and principal solution. There are an infinite number of positive and negative angles that satisfy an equation. We cannot write all of them, so we generalize them using an integer so that one can find any solution by just putting the value of integer. However, in principle there is a limit given to us between which we are required to find the solution. Here we are going to use the sum to product rule.
Now, let us come to the question

$\begin{align}
  & \sin 2x-\sin 4x+\sin 6x=0 \\
 & \sin 2x+\sin 6x-\sin 4x=0 \\
 & 2\sin \left( \dfrac{2x+6x}{2} \right)\cos \left( \dfrac{6x-2x}{2} \right)-\sin 4x=0 \\
 & 2\sin 4x\cos 2x-\sin 4x=0 \\
 & \sin 4x\left( 2\cos 2x-1 \right)=0 \\
\end{align}$
$\begin{align}
  & \sin 4x=0\text{ or 2cos 2x}-1=0 \\
 &\sin 4x=0\text{ or }\cos 2x=\dfrac{1}{2} \\
 &4x=n\pi \text{ or }2x=2n\pi \pm \dfrac{\pi }{3} \\
 &x=\dfrac{n\pi }{4}\text{ or }x=n\pi \pm \dfrac{\pi }{6} \\
\end{align}$

Note: It is important to note that if, $\sin y=0\Rightarrow y=n\pi $. If $\cos y=\cos a\Rightarrow y=2n\pi \pm a$, hence in the above solution it was in the form of $\cos y=\dfrac{1}{2}=\cos \dfrac{\pi }{3}$, so we have applied $y=2n\pi \pm \dfrac{\pi }{3}$.