Question

Solve the given equation for x, sin2x – sin4x + sin6x = 0

Answer 1

Hint: First we use the formula $\sin C+\sin D=2\sin \left( \dfrac{C+D}{2} \right)\cos \left( \dfrac{C-D}{2} \right)$ , after that we will take out the common expression and find the general solutions of the two equations separately and that will be the final answer.

Complete step-by-step answer:
Let’s start solving the question,
sin2x – sin4x + sin6x = 0
Now using the formula $\sin C+\sin D=2\sin \left( \dfrac{C+D}{2} \right)\cos \left( \dfrac{C-D}{2} \right)$ in sin2x and sin6x we get,
$\sin 2x+\sin 6x=2\sin \left( \dfrac{2x+6x}{2} \right)\cos \left( \dfrac{6x-2x}{2} \right)$
sin2x + sin6x = 2(sin4x)(cos2x)
Now using the above value in sin2x – sin4x + sin6x = 0 we get,
$\begin{aligned} & 2\sin 4x\cos 2x-\sin 4x=0 \\\ & \sin 4x\left( 2\cos 2x-1 \right)=0 \\\ \end{aligned}$
From this we can see that the there are two equations,
sin4x = 0 and 2cos2x – 1 = 0
Let’s first solve sin4x = 0,
We know that sin0 = 0,
Hence, sin4x = sin0
Now, if we have $\sin \theta =\sin \alpha$ then the general solution is:
$\theta =n\pi +{{\left( -1 \right)}^{n}}\alpha$
Now using the above formula for sin4x = sin0 we get,
$\begin{aligned} & 4x=n\pi +{{\left( -1 \right)}^{n}}0 \\\ & x=\dfrac{n\pi }{4}............(1) \\\ \end{aligned}$
Here n = integer.
Now we will find the general solution of 2cos2x – 1 = 0
$\begin{aligned} & \cos 2x=\dfrac{1}{2} \\\ & \cos 2x=\cos \dfrac{\pi }{3} \\\ \end{aligned}$
Now we will use the formula for general solution of cos,
Now, if we have $\cos \theta =\cos \alpha$ then the general solution is:
$\theta =2n\pi \pm \alpha$
Now using the above formula for 2cos2x – 1 = 0 we get,
$\begin{aligned} & 2x=2n\pi \pm \dfrac{\pi }{3} \\\ & x=n\pi \pm \dfrac{\pi }{6}............(2) \\\ \end{aligned}$
Here n = integer.
Now from equation (1) and (2) we can say that the answer is,
$x=\dfrac{n\pi }{4}\text{ or }x=n\pi \pm \dfrac{\pi }{6}$
Hence, this is the answer to this question.

Note: The trigonometric formula $\sin C+\sin D=2\sin \left( \dfrac{C+D}{2} \right)\cos \left( \dfrac{C-D}{2} \right)$ that we have used must be kept in mind. One can also take some different value of $\alpha$ in both the equations like in case of sin we can take $\pi$ instead of 0, and then can apply the same formula for the general solution, and the answer that we get will also be correct.