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Question: How do you solve \( 2{\sin ^2}x = \sin x \) over the interval of 0 to \( 2\pi \) ....

How do you solve 2sin⁡2x=sin⁡x2{\sin ^2}x = \sin x over the interval of 0 to 2π2\pi .

Explanation

Solution

Hint : Simplify the given equation to simple form and equate to angle value to get solutions.
First, we are going to rearrange the whole set of the given equations and equate it to zero, then we are going to take common if there are like terms and then we are going to equate them to zero which will give some value in terms of sinxsinx value. Then, we substitute them and find the respective sine angle which are going to be the solutions of the given equations.

Complete step-by-step answer :
First, we are going to rearrange the given equation such that we equate it to zero.
2sin⁡2x−sin⁡x=02{\sin ^2}x - \sin x = 0
We are like terms here we can take from these. Then, we get
sinx (2 sinx − 1) = 0sinx{\text{ }}\left( {2{\text{ }}sinx{\text{ }} - {\text{ }}1} \right){\text{ }} = {\text{ }}0
Now, we get two different equation from this which can be used to find the solutions, which is
sinx = 0sinx{\text{ }} = {\text{ }}0 and
(2 sinx − 1) = 0\left( {2{\text{ }}sinx{\text{ }} - {\text{ }}1} \right){\text{ = 0}}
Now, if we take sinx = 0sinx{\text{ }} = {\text{ }}0 , we can get the value of x as 0,π,2π0,\pi ,2\pi it goes on further but we have been given a specific interval of 0 to 2π2\pi .
Now, we take (2 sinx − 1) = 0\left( {2{\text{ }}sinx{\text{ }} - {\text{ }}1} \right){\text{ = 0}} , which can also be written as sin x = 12{\text{sin x = }}\dfrac{{\text{1}}}{2} , then we solve it for x for which the values are going to be
sin⁡5π6=sin⁡(π−π6)=sin⁡π6=12\sin 5\dfrac{\pi }{6} = \sin \left( {\pi - \dfrac{\pi }{6}} \right) = \sin \dfrac{\pi }{6} = \dfrac{1}{2}
Hence, we have found all the possible solutions for the given equation in the given interval.
The solutions are 0,  π,  2π,  π6  and  5π60,\;\pi ,\;2\pi ,\;\dfrac{\pi }{6}\;and\;5\dfrac{\pi }{6}
So, the correct answer is “ 0,  π,  2π,  π6  and  5π60,\;\pi ,\;2\pi ,\;\dfrac{\pi }{6}\;and\;5\dfrac{\pi }{6} ”.

Note : We have to be careful while finding out a solution, as we have been given a specific interval/range for which the solutions of the given equations should lie in and we have to always try to simplify given equations for easier solutions.