Question

Find all solutions to the equation.

7 sin^2x - 14 sin x + 2 = -5

Answer 1

I have, ...+7=0

Answer 2

ACtually, I have up to, 7*(sinx-1)(sinx-1)

Answer 3

So here, you'd want to use the zero-product rule. sin(x) = 1 at what place on a unit circle?

Answer 4

An example of the zero-product rule, in case:

Solve

x^2 - 3x + 2 = 0

We first factor:

(x - 2) (x - 1) = 0

Then we use the zero product rule to get

x - 2 = 0 or x - 1 = 0

Hence

x = 2 or x = 1

Answer 5

Okay, I got sinx=1

Answer 6

Sweet, so since you still have to solve for x, you find the places on a unit circle where sin is 1. http://www.mathsisfun.com/geometry/unit-circle.html

According to the unit circle at the bottom of the page of this link, sin(x)=1 at pi/2

Answer 7

Oh, okay.

Answer 8

Thanks.

Answer 9

Sorry about the late response, but what would the other answers be?

Answer 10

pi/2, pi/2, and 7=0 is what I got. I'm confused on the 7=0 though.

Answer 11

Can you please help, anyone?

Answer 12

What would I do with 7=0?

Answer 13

When you have an equation such as 2(x + 1)(x - 2) = 0, and you use the zero product rule, that only applies to factors that can actually equal zero. 2 = 2 and cannot equal zero, so don't do anything with it.
x + 1 = 0 or x - 2 = 0
will give you solutions
x = -1 or x = 2.

Actually, there is a mathematical thing you can do at this stage:
2(x + 1)(x - 2) = 0
Before applying the zero product rule, divide both sides by 2:
(x + 1) (x - 2) = 0
x + 1 = 0 or x - 2 = 0
x = -1 or x = 2

Your case is
7 sin^2x - 14 sin x + 2 = -5
7 sin^2x - 14 sin x + 7 = 0
7(sin^2x - 2 sin x + 1) = 0
7(sin x - 1)(sin x - 1) = 0
Now divide both sides by 7
sin x - 1 = 0 or sin x - 1 = 0
sin x = 1 or sin x = 1
You get x = pi/2 twice, and that's it.