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Question: How do you solve \({\sin ^2}x - \sin x = 0\) for \(0 \leqslant x \leqslant 2\pi \)?...

How do you solve sin2xsinx=0{\sin ^2}x - \sin x = 0 for 0x2π0 \leqslant x \leqslant 2\pi ?

Explanation

Solution

The given equation is a quadratic equation in sinx\sin x. First factorize the left hand side of the equation by taking the common term outside. Then equate both the factors to zero to determine the value of xx. Keep in mind that only those values of xx are valid which satisfy the condition 0x2π0 \leqslant x \leqslant 2\pi .

Complete step by step answer:
According to the question, we have to show the method to solve the given trigonometric equation under the provided condition.
The equation is:
sin2xsinx=0\Rightarrow {\sin ^2}x - \sin x = 0 for 0x2π0 \leqslant x \leqslant 2\pi
As we can see that this is a quadratic equation in sinx\sin x. We will solve it by factoring the quadratic equation.
Taking common term outside of the left hand side of the equation, we have:
sinx(sinx1)=0\Rightarrow \sin x\left( {\sin x - 1} \right) = 0
Thus sinx\sin x and sinx1\sin x - 1 are two factors of the expression. Since the right hand side is zero, we’ll equate both these factors to zero. This we’ll get:
sinx=0 and (sinx1)=0 sinx=0 and sinx=1  \Rightarrow \sin x = 0{\text{ and }}\left( {\sin x - 1} \right) = 0 \\\ \Rightarrow \sin x = 0{\text{ and }}\sin x = 1 \\\
Case 1:
We’ll find the values of xx such that sinx=0\sin x = 0 and 0x2π0 \leqslant x \leqslant 2\pi . We already know that the value of sinx\sin x is zero for integral multiples of π\pi . Thus the values of xx in the given range are:
x=0, π and 2π\Rightarrow x = 0,{\text{ }}\pi {\text{ and }}2\pi
Case 2:
Similarly we’ll find the values of xx such that sinx=1\sin x = 1 and 0x2π0 \leqslant x \leqslant 2\pi . We know that the value of sinx\sin x is 1 for x=nπ2x = \dfrac{{n\pi }}{2} where n=1,5,9,...n = 1,5,9,.... Thus the only value of xx in the given range is:
x=π2\Rightarrow x = \dfrac{\pi }{2}
Hence if we combine the values of both the cases, the values of xx are:
x=0, π2π and 2π\Rightarrow x = 0,{\text{ }}\dfrac{\pi }{2}{\text{, }}\pi {\text{ and }}2\pi
Therefore these are the solutions of the equation.

Note: sinx\sin x is a periodic function and its period is 2π2\pi . If we understand the values and behaviour of the function within 0x2π0 \leqslant x \leqslant 2\pi then it will be repeated before and after this interval of xx. The range of sinx\sin x is from -1 to 1 i.e. R[1,1]R \in \left[ { - 1,1} \right]. Its value is 1 and -1 only once within 0x2π0 \leqslant x \leqslant 2\pi (i.e. at x=π2x = \dfrac{\pi }{2} and x=3π2x = \dfrac{{3\pi }}{2} respectively) and its value is 0 thrice within 0x2π0 \leqslant x \leqslant 2\pi (i.e. at x=0, π and 2πx = 0,{\text{ }}\pi {\text{ and }}2\pi ). All other values from [1,1]\left[ { - 1,1} \right] occur twice in this interval of xx.