Question
Question: How do you solve \({\sin ^2}x - \sin x = 0\) for \(0 \leqslant x \leqslant 2\pi \)?...
How do you solve sin2x−sinx=0 for 0⩽x⩽2π?
Solution
The given equation is a quadratic equation in sinx. 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⩽x⩽2π.
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:
⇒sin2x−sinx=0 for 0⩽x⩽2π
As we can see that this is a quadratic equation in sinx. We will solve it by factoring the quadratic equation.
Taking common term outside of the left hand side of the equation, we have:
⇒sinx(sinx−1)=0
Thus sinx and sinx−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 (sinx−1)=0 ⇒sinx=0 and sinx=1
Case 1:
We’ll find the values of x such that sinx=0 and 0⩽x⩽2π. We already know that the value of sinx is zero for integral multiples of π. Thus the values of x in the given range are:
⇒x=0, π and 2π
Case 2:
Similarly we’ll find the values of x such that sinx=1 and 0⩽x⩽2π. We know that the value of sinx is 1 for x=2nπ where n=1,5,9,.... Thus the only value of x in the given range is:
⇒x=2π
Hence if we combine the values of both the cases, the values of x are:
⇒x=0, 2π, π and 2π
Therefore these are the solutions of the equation.
Note: sinx is a periodic function and its period is 2π. If we understand the values and behaviour of the function within 0⩽x⩽2π then it will be repeated before and after this interval of x. The range of sinx is from -1 to 1 i.e. R∈[−1,1]. Its value is 1 and -1 only once within 0⩽x⩽2π (i.e. at x=2π and x=23π respectively) and its value is 0 thrice within 0⩽x⩽2π (i.e. at x=0, π and 2π). All other values from [−1,1] occur twice in this interval of x.