1. ## Find all solutions

$\displaystyle -cos^2x- 4 sin \frac{1}{2}x+2=0$

2. Originally Posted by purplec16
$\displaystyle -cos^2x- 4 sin \frac{1}{2}x+2=0$
note that this is not an easy equation to solve by hand.

(sure that you posted it correctly?)

If correct, then I recommend you graph the left side of the equation over the interval from $\displaystyle 0$ to $\displaystyle 2\pi$ and locate the zeros using your calculator.

3. Originally Posted by purplec16
$\displaystyle -cos^2x- 4 sin \frac{1}{2}x+2=0$
The sine of the half-angle formula should (?) help here:

sin(x/2)=+- sqrt[(1−cosx)/2]

Good luck!

4. Hello purplec16
Originally Posted by purplec16
$\displaystyle -cos^2x- 4 sin \frac{1}{2}x+2=0$
If you're going to try for an analytical solution, you can reduce this equation to a quartic in $\displaystyle s$, where $\displaystyle s = \sin\tfrac12x$, using $\displaystyle \cos x = 1 - 2\sin^2x$. The equation then becomes:
$\displaystyle 4s^4 -4s^2+4s-1=0$
but, apart from adopting a numerical/graphical method (as skeeter suggested), I can't see what to do next.