Question

How do you solve $\sin 2x = \sin x$ over the interval $0$ to $2\pi$?

Answer 1

In this problem, we have to solve the equation which contains trigonometric functions. We have to find the values of $x$ in the interval $\left[ {0,2\pi } \right]$ such that $\sin 2x = \sin x$. To solve the given equation, we will use the trigonometric formula which is given by $\sin 2\theta = 2\sin \theta \cos \theta$. Also we must know some trigonometric values.

Complete step-by-step answer:
In this problem, the given equation is $\sin 2x = \sin x \cdots \cdots \left( 1 \right)$. Solve this equation over the interval $\left[ {0,2\pi } \right]$ means we have to find all values of $x$ in the interval $\left[ {0,2\pi } \right]$ which satisfy the equation $\left( 1 \right)$.
We know the trigonometric formula $\sin 2x = 2\sin x\cos x$. Use this formula on the left-hand side of the equation $\left( 1 \right)$. So, we can write $2\sin x\cos x = \sin x \cdots \cdots \left( 2 \right)$.
Let us rewrite the equation $\left( 2 \right)$ by taking the right-hand side term on the left-hand side. So, we get $2\sin x\cos x - \sin x = 0 \cdots \cdots \left( 3 \right)$. Taking equal term $\sin x$ common out from equation $\left( 3 \right)$, we get $\sin x\left( {2\cos x - 1} \right) = 0 \cdots \cdots \left( 4 \right)$.
We know that if $ab = 0$ then either $a = 0$ or $b = 0$. Use this information in the equation $\left( 4 \right)$, we can write $\sin x = 0 \cdots \cdots \left( 5 \right)$ or $2\cos x - 1 = 0 \cdots \cdots \left( 6 \right)$.
To solve the equation $\left( 5 \right)$, we have to think about the values of $x$ in the interval $\left[ {0,2\pi } \right]$ such that $\sin x = 0$. We know that $\sin 0 = 0,\;\sin \pi = 0,\;\sin 2\pi = 0$. Hence, we can write
$\sin x = 0 \Rightarrow x = 0,\pi ,2\pi$
To solve the equation $\left( 6 \right)$, first we will rewrite the equation. So, we get $2\cos x - 1 = 0 \Rightarrow 2\cos x = 1 \Rightarrow \cos x = \dfrac{1}{2}$. Now we have to think about the values of $x$ in the interval $\left[ {0,2\pi } \right]$ such that $\cos x = \dfrac{1}{2}$. We know that $\cos \dfrac{\pi }{3} = \dfrac{1}{2},\;\cos \dfrac{{5\pi }}{3} = \dfrac{1}{2}$. Hence, we can write
$\cos x = \dfrac{1}{2} \Rightarrow x = \dfrac{\pi }{3},\dfrac{{5\pi }}{3}$.

Hence, the solution set of the given equation is \left\\{ {0,\dfrac{\pi }{3},\pi ,\dfrac{{5\pi }}{3},2\pi } \right\\}.

Note:
When the trigonometric functions are involved in the given problem then trigonometric identities (formulas) and trigonometric values are very useful to solve the problem. In this problem, if the interval $\left[ {0,2\pi } \right]$ is not given then we can find infinitely many solutions.