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?
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?
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.
Hello, sinewave85!
Have you ever tried to factor something with a square root in it?
$\displaystyle -4\cos^2\!x + 2\sqrt{1-\cos^2x} + 3 \;=\;0$
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 . . . ]
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.