# Solving Trig Ids

• Jan 24th 2010, 07:14 AM
TeriyakiDonnQ
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 = 2-sinx
then changed it the cos^2 to.. 2( 1-sin^2x) = 2-sinx

Am i doing these right in the first place..?
• Jan 24th 2010, 07:58 AM
craig
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:

$\displaystyle S^2 - \frac{1}{2}S = 0$
• Jan 24th 2010, 08:06 AM
TeriyakiDonnQ
ohh~ okayy :D
i get it thank youuu! (Clapping)