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

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- Mar 17th 2010, 07:17 AMpurplec16Find all solutions
$\displaystyle -cos^2x- 4 sin \frac{1}{2}x+2=0$

- Mar 17th 2010, 08:52 AMskeeter
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. - Mar 17th 2010, 08:54 AMapcalculus
- Mar 17th 2010, 11:49 AMGrandad
Hello purplec16If 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.

Grandad