Solve the equation for the interval [0, 2pi). 2sin^2x = sin x A) 0, pi, pi/6, 5pi/6
B) pi/6, 5pi/6
C) pi/3, 2pi/3
D) pi/2, 3pi/2, pi/3, 2pi/3​

A

Step-by-step explanation:

We want to solve the equation:

For the interval [0, 2π).

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

We can factor:

By the Zero Product Property:

Solve for each case:

Use the unit circle to solve:

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.

a

step-by-step explanation:

ca = ec

cpctc makes it so that corresponding parts of the triangle you proved to be equal are equal, the other options aren't part of the part of the triangles you proved to be equal.