Question

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

Answer 1

Hint: The given equation is a quadratic equation in $\sin x$. 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 $x$. Keep in mind that only those values of $x$ are valid which satisfy the condition $0 \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:
$ \Rightarrow {\sin ^2}x - \sin x = 0$ for $0 \leqslant x \leqslant 2\pi $
As we can see that this is a quadratic equation in $\sin x$. We will solve it by factoring the quadratic equation.
Taking common term outside of the left hand side of the equation, we have:
\[ \Rightarrow \sin x\left( {\sin x - 1} \right) = 0\]
Thus \[\sin x\] and \[\sin x - 1\] 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:
$
   \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 $x$ such that $\sin x = 0$ and $0 \leqslant x \leqslant 2\pi $. We already know that the value of \[\sin x\] is zero for integral multiples of $\pi $. Thus the values of $x$ in the given range are:
$ \Rightarrow x = 0,{\text{ }}\pi {\text{ and }}2\pi $
Case 2:
Similarly we’ll find the values of $x$ such that $\sin x = 1$ and $0 \leqslant x \leqslant 2\pi $. We know that the value of \[\sin x\] is 1 for $x = \dfrac{{n\pi }}{2}$ where $n = 1,5,9,...$. Thus the only value of $x$ in the given range is:
$ \Rightarrow x = \dfrac{\pi }{2}$
Hence if we combine the values of both the cases, the values of $x$ are:
$ \Rightarrow x = 0,{\text{ }}\dfrac{\pi }{2}{\text{, }}\pi {\text{ and }}2\pi $
Therefore these are the solutions of the equation.

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