Ok, working on a trig equation, have part of it down to:

$\displaystyle -4\cos^{2}x+2\sqrt{1-\cos^{2}x}+3=0$

I need the left side factored (so I can use the zero rule to find values for x) but I can't seem to get it. Can anyone else?

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- Nov 14th 2006, 12:59 PMsinewave85need factors part of trig eqation
Ok, working on a trig equation, have part of it down to:

$\displaystyle -4\cos^{2}x+2\sqrt{1-\cos^{2}x}+3=0$

I need the left side factored (so I can use the zero rule to find values for x) but I can't seem to get it. Can anyone else? - Nov 14th 2006, 01:19 PMThePerfectHacker
Note, the error $\displaystyle \sqrt{1-\cos^2 x}=\sin x$

In fact it is, $\displaystyle \sqrt{1-\cos^2 x}=|\sin x|$

The presense of the absolute value function overcomplicates things.

What you can do, is express it as, (add and subtract 1)

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

Thus,

$\displaystyle 4(1-\cos^2 x)+2\sqrt{1-\cos^2 x}-1=0$

If,

$\displaystyle \sqrt{1-\cos^2 x}=y$

Then,

$\displaystyle 4y^2+2y-1=0$

You have a quadradic equation.

Just remember to check solutions because it is necessary though not suffienct. - Nov 14th 2006, 01:28 PMSoroban
Hello, sinewave85!

Have you**ever**tried to factor something with a square root in it?

Quote:

$\displaystyle -4\cos^2\!x + 2\sqrt{1-\cos^2x} + 3 \;=\;0$

Remember "radical equations" from Algebra 2?

We have: .$\displaystyle 2\sqrt{1-\cos^2\!x} \;=\;4\cos^2\!x - 3$

Square both sides: .$\displaystyle 4\left(1 - \cos^2\!x\right) \;= \; 16\cos^4\!x - 24\cos^2\!x + 9$

. . and we have: .$\displaystyle 16\cos^4\!x - 20\cos^2\!x + 5\;=\;0$

Quadratic Formula: .$\displaystyle \cos^2\!x \;= \;\frac{-(-20) \pm \sqrt{20^2 - 4(16)(5)}}{2(16)} \;=\;\frac{20 \pm\sqrt{80}}{32}$

. . and we have: .$\displaystyle \cos^2\!x \;=\;\frac{5 \pm\sqrt{5}}{8}\quad\Rightarrow\quad\boxed{\cos x\;=\;\pm\sqrt{\frac{5\pm\sqrt{5}}{8}}} $

Don't forget to check for extraneous roots.

[And you wonder why you couldn't factor it . . . ] - Nov 14th 2006, 02:11 PMsinewave85
Thanks for the help -- both of you! I got it worked out, and I got the right answers, along with a few false ones. Just in case you are currious, the original problem was sin2x = cos3x, and the answers for x between 0 and 180 degrees are 18, 90, and 160 degrees. I am not sure I solved it the best possible way, but I did get the right answers -- with a little help ;). Thanks again for getting me through my brain fart.