
Solving Trig Ids
I'm given
cosx (tanx) = 0
and
2cos^2x + sin x = 2
The first one i did...
cosx ( sinx / cosx ) = 0
then i got rid of the cosx from sinx / cosx and got...
cos^2x ( sinx ) = 0
is that even right? D: Am I just solving for x or what? D:
for 2cos^2x + sin x = 2 ...
i moved sinx over to the 2... 2cos^2x = 2sinx
then changed it the cos^2 to.. 2( 1sin^2x) = 2sinx
Am i doing these right in the first place..?

Quote:
Originally Posted by
TeriyakiDonnQ cosx (tanx) = 0
You started right with this.
$\displaystyle cos(x)(tan(x)) = cos(x)(\frac{sin(x)}{cos(x)})$
This is the same as $\displaystyle \frac{cos(x)sin(x)}{cos(x)} = sin(x) = 0$.
You then solve this for x.
For the second one again you had the right idea to start with, let $\displaystyle cos^2(x) = 1  sin^2(x)$, your equation is now:
$\displaystyle 2(1  sin^2(x)) + sin(x)2 = 0$
Rearranging this gives us $\displaystyle sin^2(x)  \frac{1}{2}sin(x) = 0$
If we now replace $\displaystyle sin(x)$ with $\displaystyle S$, we get a regular quadratic in $\displaystyle S$ to solve:
$\displaystyle S^2  \frac{1}{2}S = 0$

ohh~ okayy :D
i get it thank youuu! (Clapping)