Question

Consider the following trigonometric equation:
$\sin x+\sin 3x+\sin 5x=0$, then the solution is:
A. $x=\dfrac{2n\pi }{3},n\in I$
B. $\dfrac{n\pi }{3},n\in I$
C. $\left( 3n\pm 1 \right)\dfrac{\pi }{3},n\in I$
D. none of these

Answer 1

Hint: We will use the trigonometric identity to simplify the expression as shown below:
$\sin C+\sin D=2\sin \left( \dfrac{C+D}{2} \right)\cos \left( \dfrac{C-D}{2} \right)$
Also, we will use the general solution for $\sin x=\sin \alpha$ given by $x=n\pi +{{\left( -1 \right)}^{n}}\alpha$ and for $\cos x=\cos \alpha$ is $x=2n\pi \pm \alpha$ where $n\in I$.

Complete step-by-step answer:
We have been given $\sin x+\sin 3x+\sin 5x=0$
On rearranging the terms, we get as follows:
$\Rightarrow \sin x+\sin 5x+\sin 3x=0$
We know that $\sin C+\sin D=2\sin \left( \dfrac{C+D}{2} \right)\cos \left( \dfrac{C-D}{2} \right)$

& \Rightarrow 2\sin \left( \dfrac{x+5x}{2} \right)\cos \left( \dfrac{5x-x}{2} \right)+\sin 3x=0 \\\ & \Rightarrow 2\sin \dfrac{6x}{2}\cos \dfrac{4x}{2}+\sin 3x=0 \\\ & \Rightarrow 2\sin 3x\cos 3x+\sin 3x=0 \\\ \end{aligned}$$ Taking sin3x as common, we get as follows: $$\sin 3x\left( 2\cos 2x+1 \right)=0$$ $$\Rightarrow \sin 3x=0$$ and $$2\cos 2x+1=0$$ Since we know that the general solution for $$\sin x=\sin \alpha $$ is given by $$x=n\pi +{{\left( -1 \right)}^{n}}\alpha $$ where $$n\in I$$. So, for $$\sin 3x=0$$ $$\Rightarrow \sin 3x=\sin {{0}^{\circ }}$$ $$\begin{aligned} & \Rightarrow 3x=n\pi \\\ & \Rightarrow x=\dfrac{n\pi }{3} \\\ \end{aligned}$$ Where $$n\in I$$ Also, we know that the general solution for $$\cos x=\cos \alpha $$ is given by $$x=2n\pi \pm \alpha $$. So for $$2\cos 2x+1=0$$ $$\begin{aligned} & \Rightarrow 2\cos 2x=-1 \\\ & \Rightarrow \cos 2x=\dfrac{-1}{2} \\\ & \Rightarrow \cos 2x=cos\left( \dfrac{2\pi }{3} \right) \\\ \end{aligned}$$ Since we know that $$\cos \left( \dfrac{2\pi }{3} \right)=\dfrac{-1}{2}$$ $$\begin{aligned} & \Rightarrow 2x=2n\pi \pm \dfrac{2\pi }{3} \\\ & \Rightarrow x=n\pi \pm \dfrac{\pi }{3} \\\ & \Rightarrow x=\left( 3n\pm 1 \right)\dfrac{\pi }{3} \\\ \end{aligned}$$ Where $$n\in I$$ Hence, the solution of the given equation is $$x=\dfrac{n\pi }{3}$$ and $$x=\left( 3n\pm 1 \right)\dfrac{\pi }{3}$$, where $$n\in I$$ (integers). Therefore, the correct answers are options B and C. Note: Be careful while using the trigonometric identity to simplify the given expression and take care of the sign as there is a chance that you might make a sign mistake. We must be aware while using the value of $$\cos \dfrac{\pi }{3}=\dfrac{1}{2}$$ and $$\cos \dfrac{2\pi }{3}=\dfrac{-1}{2}$$. The negative sign should be taken into consideration. Also, remember that there are two solutions that we would obtain. Hence, two options will be correct. Sometimes we just forget to solve the other part of the solution and we get an incomplete solution for the equation.