Question

How do you solve the equation $\sin 2x-\sin x=0$ on the interval $[0,2\pi ]$?

Answer 1

Hint:The given equation $\sin 2x-\sin x=0$is a linear equation in $1$ variable since this is a first-order equation. Linear equations are solved for the value of an unknown variable. In this question, $x$ is the unknown variable and we have to find the value of $x$ for all cases. The given trigonometric equation is solved using the trigonometric identities. In this question, we use the identity $\sin 2x=2\sin x\cos x$ and then continue to solve the equation. The value of x will lie between $[0,2\pi ]$ since the domain is $[0,2\pi ]$. Also, we need to take care of the sign of the trigonometry in the given interval $[0,2\pi ]$. Sin is positive in the first and second quadrant and cos is positive in the first and third quadrant.

Complete step by step solution:
We have to solve the equation $\sin 2x-\sin x=0$ in the interval $[0,2\pi ]$
Using the trigonometric identity $\sin 2x=2\sin x\cos x$ in the given equation, we get
$\begin{align}
  & 2\sin x\cos x-\sin x=0 \\
 & \Rightarrow \sin x(2\cos x-1)=0 \\
\end{align}$
either $\sin x=0$ or $2\cos x-1=0$
We know that,
$\sin x=0$ when $x=0,\pi ,2\pi ,...$
  But we have to solve the equation in the interval $[0,2\pi ]$.
Therefore, $x=0,\pi ,2\pi $ since $0,\pi ,2\pi \in [0,2\pi ]$
Now,
 $2\cos x-1=0$
  $
 \Rightarrow 2\cos x=1 \\
 \Rightarrow \cos x=1/2 \\
$
We know that,
 $\cos x=1/2$ when $x=\pi /3,\pi +\pi /3,2\pi +\pi /3,...$
$\cos x=1/2$ when $x=\pi /3,4\pi /3,7\pi /3,...$
But we have to find the solution in the interval $[0,2\pi ]$.
Therefore, $x=\pi /3,4\pi /3$ since $\pi /3,4\pi /3\in [0,2\pi ]$

Therefore, the solution of the equation $\sin 2x-\sin x=0$ is:
$x=0,\pi ,2\pi ,\pi /3,4\pi /3$
And x lies in the interval $[0,2\pi ]$.

Note: $\sin x=0$ when $x=0,\pi ,2\pi ,...$ i.e. when $x=\pm nx$, where $n=0,1,2,...$ and $\cos x=1/2$ is only possible in the first and third quadrant where cos gives positive value . In the first quadrant, value of $\cos \pi /3=1/2$ and in the third quadrant, the value of $\cos (\pi +\pi /3)=1/2 \cos 4\pi /3=1/2$. Similarly, $\cos (2\pi +\pi /3)=1/2$ and so on. That is, $\cos x=1/2$ when $x=\pi /3,\pi +\pi /3,2\pi +\pi /3,...$ i.e. $\cos x=1/2$ when $x=n\pi +\pi /3$, where $n=0,1,2,...$ . There are a lot of trigonometric identities which we can use to solve the trigonometric equations.