# Homework Help: Trig equations

1. Jul 26, 2011

### Nelo

1. The problem statement, all variables and given/known data

I dont understand how to solve for these, I will post 3.

d) [sqrt]2sin x + 1 = 0 (sqrt only over the 2)
e) 2cos x - [sqrt]3 = 0
f) 2sinx +[sqr]3 = 0

Esentially, I dont understand how to get the degrees. I dont get how you use the terminal arm to predict the degrees, etc.

Can anyone help?

2. Relevant equations

3. The attempt at a solution

1 / [sqrt]2

sin inverse of that is 30 degrees.

which is pi over 6.

How do i get the second one? how do i know? etc

2. Jul 26, 2011

### rock.freak667

I am assuming you are finding from 0 to 2pi.

For the three you are doing the same thing, making sinx or cosx the subject of the formula and then taking the inverse function of it.

The first one you did partially right

sinx = 1/√2 and x=sin-1(1/√2)

Check back on your calculator for what value this is.

When you get that what quadrant is sine positive in?

3. Jul 27, 2011

### Ray Vickson

Write sqrt(2) instead of [sqrt]2, etc; that way, you don't have to make statements like "sqrt only over the 2". Anyway, if you know elementary Geometry, you should be able to see what angles give you sin(x) = -1/sqrt(2), cos(x) = 3/sqrt(2) and sin(x) = -sqrt(3)/2. I realize that Geometry may not be taught anymore in schools, in which case you can do a Google search on something like "sines of special angles" to find some material.

RGV

4. Aug 2, 2011

### Nelo

4f) cosx =2sincosx

How do you solve this?

The only thing i see is

..cosx - cosx- = 2sin

is that wrong?

does that mean cos is 0 and sin is 90+90 = 180 degrees?

5. Aug 2, 2011

### rock.freak667

bring the cosx on the left side to the right side and then factor out the cosx.

6. Aug 2, 2011

### Nelo

Does that not equal the exact same thing? cos is 0 and sin is 180?

7. Aug 2, 2011

### rock.freak667

You mean x=0, x=180? if so, by putting x=0 into the equation you will see that it is not a solution.