1. ## Quadrant solutions

2 sin 4x + sqrt3 = 0

Rewritten:
2 sin 4x = - sqrt3

sqrt3 can be a positive or negative number so there are solutions in all quadrants? My book says only the third and fourth. I don't need any further work done.

2. Originally Posted by Stuck Man
2 sin 4x + sqrt3 = 0

Rewritten:
2 sin 4x = - sqrt3

sqrt3 can be a positive or negative number so there are solutions in all quadrants? My book says only the third and fourth. I don't need any further work done.
hi

your book is right . $\sin 4x=-\frac{\sqrt{3}}{2}$ and sin is negative in teh 3rd and 4th quadrant .

Note that $x=\sqrt{4}\Rightarrow x=2$ but if $x^2=4\Rightarrow x=\pm 2$

3. I didn't know that. I've been making that mistake for a while now.