@tkhunny

Solve to four decimal places:

sin^2 x + 2 cos x=-2

0

Question

@tkhunny Solve to four decimal places: sin^2 x + 2 cos x=-2 0

tkhunny · Answer

Use the Pythagorean Identity.  You'll be done in no time.

$sin^{2}(x) = 1 - \cos^{2}(x)$

anonymous · Answer

I get the feeling I'm doing something very wrong, I'm down to determining wether the answer is -0.25+1.1975i and -0.25-1.1975i

tkhunny · Answer

You must show your work.  Just dropping the answer is not useful.  No one can tell what you are doing.

$\sin^{2}(x) + 2\cos(x)=-2 $

$1 - \cos^{2}(x) + 2\cos(x)=-2 $

$- \cos^{2}(x) + 2\cos(x) + 3 = 0$

$\cos^{2}(x) - 2\cos(x) - 3 = 0$

$(\cos(x) - 3)(\cos(x)+1) = 0$

$\cos(x) - 3 = 0 \implies \cos(x) = 3$ -- Well, That's just silly.  No solution from that.
$\cos(x) + 1 = 0 \implies \cos(x) = -1$ -- That's good.  x = 180º

Stare at this for a very long time.  EVERYTHING I did is there for your viewing pleasure.  You can follow it.  You can reproduce it.  If there are errors, you can find them and fix them.  THAT is what "show your work" means.

anonymous · Answer

I did manage to find my mistakes and completed and sent in my homework, thanks for your help!!

tkhunny · Answer

Well, that's good, I suppose, but if you really want useful assistance, you will have to show your work to those from whom you seek said assistance.  Good work.