Question

How do you find all values of x such that $\sin 2x=\sin x$ and $0\le x\le 2\pi $?

Answer 1

Hint: In the above question, we have been given a trigonometric equation, which is written as $\sin 2x=\sin x$. For solving it, we can apply the trigonometric identity given by $\sin 2x=2\sin x\cos x$ so that the equation will become $2\sin x\cos x=\sin x$. Then, we need to subtract $\sin x$ from both the sides to get $\sin x\left( 2\cos x-1 \right)=0$. From this, we will obtain two equations. On will be $\sin x=0$ and the other will be $\cos x=\dfrac{1}{2}$. Using the principle solutions of these equations, and according to the given interval $0\le x\le 2\pi $, we will get all the solutions of the given equations.

Complete step-by-step answer:
The trigonometric equation given in the question is
$\Rightarrow \sin 2x=\sin x$
We know that $\sin 2x=2\sin x\cos x$. Substituting this in the above equation, we get
$\Rightarrow 2\sin x\cos x=\sin x$
Now, on subtracting $\sin x$ from both the sides of the above equation, we get
\[\begin{align}
  & \Rightarrow 2\sin x\cos x-\sin x=\sin x-\sin x \\
 & \Rightarrow \sin x\left( 2\cos x-1 \right)=0 \\
\end{align}\]
Using the zero product rule, we get
$\begin{align}
  & \Rightarrow \sin x=0,2\cos x-1=0 \\
 & \Rightarrow \sin x=0,\cos x=\dfrac{1}{2} \\
\end{align}$
Considering the first equation, we have
$\Rightarrow \sin x=0$
In the given interval $0\le x\le 2\pi $, we get the solutions as $x=0,\pi ,2\pi $.
Now, considering the second equation, we have
$\Rightarrow \cos x=\dfrac{1}{2}$
We know that the principle solution of this equation is $x=\dfrac{\pi }{3}$. In the given interval $0\le x\le 2\pi $, all the four quadrants are involved. The solution $x=\dfrac{\pi }{3}$ belongs to the first quadrant. We know that cosine is positive in the first and the fourth quadrants. Therefore, the solution in the fourth quadrant will be \[x=2\pi -\dfrac{\pi }{3}=\dfrac{5\pi }{3}\].
Hence, the solutions of the given trigonometric equation are $x=0,x=\pi ,x=\dfrac{\pi }{3},x=\dfrac{5\pi }{3},x=2\pi $.

Note: Do not cancel $\sin x$ from the LHS and the RHS in the equation $2\sin x\cos x=\sin x$. This is because we cannot simply cancel a variable term from an equation. Also, we must not ignore the interval given in the question. All the solutions must be within the given interval.