Trig equations

[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

[rock.freak667]

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

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

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

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?

[Ray Vickson]

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

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

[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?

[rock.freak667]

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?

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

[Nelo]

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

[rock.freak667]

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

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