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Question: How do you find all values of x such that \(\sin 2x=\sin x\) and \(0\le x\le 2\pi \)?...

How do you find all values of x such that sin⁡2x=sin⁡x\sin 2x=\sin x and 0≤x≤2π0\le x\le 2\pi ?

Explanation

Solution

In the above question, we have been given a trigonometric equation, which is written as sin⁡2x=sin⁡x\sin 2x=\sin x. For solving it, we can apply the trigonometric identity given by sin⁡2x=2sin⁡xcos⁡x\sin 2x=2\sin x\cos x so that the equation will become 2sin⁡xcos⁡x=sin⁡x2\sin x\cos x=\sin x. Then, we need to subtract sin⁡x\sin x from both the sides to get sin⁡x(2cos⁡x−1)=0\sin x\left( 2\cos x-1 \right)=0. From this, we will obtain two equations. On will be sin⁡x=0\sin x=0 and the other will be cos⁡x=12\cos x=\dfrac{1}{2}. Using the principle solutions of these equations, and according to the given interval 0≤x≤2π0\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
⇒sin⁡2x=sin⁡x\Rightarrow \sin 2x=\sin x
We know that sin⁡2x=2sin⁡xcos⁡x\sin 2x=2\sin x\cos x. Substituting this in the above equation, we get
⇒2sin⁡xcos⁡x=sin⁡x\Rightarrow 2\sin x\cos x=\sin x
Now, on subtracting sin⁡x\sin x from both the sides of the above equation, we get

& \Rightarrow 2\sin x\cos x-\sin x=\sin x-\sin x \\\ & \Rightarrow \sin x\left( 2\cos x-1 \right)=0 \\\ \end{aligned}$$ Using the zero product rule, we get $\begin{aligned} & \Rightarrow \sin x=0,2\cos x-1=0 \\\ & \Rightarrow \sin x=0,\cos x=\dfrac{1}{2} \\\ \end{aligned}$ Considering the first equation, we have $\Rightarrow \sin x=0$ In the given interval $0\le x\le 2\pi $, we get the solutions as $x=0,\pi ,2\pi $. Now, considering the second equation, we have $\Rightarrow \cos x=\dfrac{1}{2}$ We know that the principle solution of this equation is $x=\dfrac{\pi }{3}$. In the given interval $0\le x\le 2\pi $, all the four quadrants are involved. The solution $x=\dfrac{\pi }{3}$ belongs to the first quadrant. We know that cosine is positive in the first and the fourth quadrants. Therefore, the solution in the fourth quadrant will be $$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.