Hello Captcha Originally Posted by

**Captcha** 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$\displaystyle \sin^2 x = \frac12$

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

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

Grandad