1. ## Trig Help.

I need help with the following equations to solve between 0 and 360°:

1. $\displaystyle \sin2x = \frac{1}{2}$

2. $\displaystyle \sin^2x = \frac{1}{2}$

3. $\displaystyle \cos(x-30) = 1$

4. $\displaystyle \cos^2x = \frac{3}{2}$

For 1, I've obtained x = 45°, 225°, 315°, 360°.
For 2, I've 30° and 75°.
For 3, hopeless.
For 4, no idea.

2. sin(2x)=1/2
let b=2x
values of b that give +1/2 are 30 and 150
so sinb=1/2 where b =30,150
b=2x implies 2x=30 and 2x = 150

sin^2 x =1/2 is the same as (sinx)^1/2 =1/2,
thus sinx = sqrt(1/2)
values of x that give +1/sqrt(2) are 45 and 135

cos(x-30)=1
let x-30=b
so cosb=1
values of b that give +1 are 0 and 360
this means x-30=0 and x-30=360

cos^2 x =3/2
this is the same as cosx=sqrt(3/2)
but sqrt(3/2) is > than 1.
cosx has a range of 1>y>-1
so there is no solution

cos^2 x =3/2
this is the same as cosx=sqrt(3/2)
but sqrt(3/2) is > than 1.
cosx has a range of 1>y>-1
so there is no solution
Hi, thanks. What about if it was $\displaystyle \cos^2x = \frac{3}{4}$? Would it have solutions?

4. Got it now! Can I thank myself? lol.

I need help with the following equations to solve between 0 and 360°:

1. $\displaystyle \sin2x = \frac{1}{2}$

2. $\displaystyle \sin^2x = \frac{1}{2}$

...
I have just a couple of things to add to questions 1 and 2.

1. If $\displaystyle 0^o\le x \le 360^o$, then $\displaystyle 0^o\le 2x \le 720^o$. So you will need to say
$\displaystyle 2x = 30^o, 150^o, 390^o, 510^o$

$\displaystyle \Rightarrow x = 15^o, 75^o, 195^o, 255^o$
2. Don't forget the $\displaystyle \pm$ sign when you take the square root. So
$\displaystyle \sin^2 x = \frac12$

$\displaystyle \Rightarrow \sin x = \pm \frac{1}{\sqrt2}$

$\displaystyle \Rightarrow x = 45^o, 135^o, 225^o, 315^o$