Question

Solve the equation for the interval [0, 2pi). 2sin^2x = sin x

Answer 1

Answer:

A

Step-by-step explanation:

We want to solve the equation:

$2\sin^2(x)=\sin(x)$

For the interval [0, 2π).

First, we can move all the terms to one side. Start off by subtracting sin(x) from both sides:

$2\sin^2(x)-\sin(x)=0$

We can factor:

$\sin(x)\left(2\sin(x)-1\right)=0$

By the Zero Product Property:

$\sin(x)=0\text{ or } 2\sin(x)-1=0$

Solve for each case:

$\displaystyle \sin(x)=0\text{ or } \sin(x)=\frac{1}{2}$

Use the unit circle to solve:

$\displaystyle x=\left\{0, \frac{\pi}{6},\frac{5\pi}{6}, \pi\right\}$

Hence, our answer is A.

*Please note that we should not simply divide both sides by sin(x) to acquire 2sin(x) = 1. The problem with the operation is that we are dividing by sin(x), yet we do not know what the value of x is. Thus, one or more values of x may result in sin(x) = 0, and we cannot divide by 0. Hence, we are required to subtract and then factor, unless the question specifically states that sin(x) ≠ 0.