1. ## Trig Equation

Given the equation $\displaystyle 2\cos^2x-\cos x-1=0$ find the exact values of x on the interval $\displaystyle [0,2\pi)$

well i tried to take out a $\displaystyle \cos x$ to come up with $\displaystyle \cos(\cos^2x-1)$ and i know that the Pythagorean identity is$\displaystyle \sin^2x=1-\cos^2x$ could anyone help steer me in the right direction

2. Originally Posted by ~berserk
Given the equation $\displaystyle 2\cos^2x-\cos x-1=0$ find the exact values of x on the interval $\displaystyle [0,2\pi)$

well i tried to take out a $\displaystyle \cos x$ to come up with $\displaystyle \cos(\cos^2x-1)$ and i know that the Pythagorean identity is$\displaystyle \sin^2x=1-\cos^2x$ could anyone help steer me in the right direction
Hi bers,

this is a quadratic from one viewpoint.

$\displaystyle cosx=y$

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

factor

$\displaystyle (2y+1)(y-1)=0$

3. I have another one: $\displaystyle 3\tan^2(2x)=1$ on the same interval. I'm not very good at getting these started because I need to become better at recognizing identities if they are in the problem.

4. Originally Posted by ~berserk
I have another one: $\displaystyle 3\tan^2(2x)=1$ on the same interval. I'm not very good at getting these started because I need to become better at recognizing identities if they are in the problem.
Sometimes you don't need any identity.
You've just got to regenerate the angles..

$\displaystyle 3tan^2(2x)=1$

$\displaystyle tan^2(2x)=\frac{1}{3}$

$\displaystyle tan2x=\pm\frac{1}{\sqrt{3}}$

Find the angles 2x up to $\displaystyle 4{\pi}$

then find x

5. Well at least it's good to know i was heading in the right direction with it before hand, it's just good to have reassurance. Would i need to use the double angle identity at all:$\displaystyle \tan 2x=\frac{2\tan x}{1-\tan^2x}$ and if so how would i implement it.

6. Originally Posted by ~berserk
Well at least it's good to know i was heading in the right direction with it before hand, it's just good to have reassurance. Would i need to use the double angle identity at all:$\displaystyle \tan 2x=\frac{2\tan x}{1-\tan^2x}$ and if so how would i implement it.
Yes, you could and you'd get another quadratic...

$\displaystyle tan2x=\frac{2tanx}{1-tan^2x}\Rightarrow\ 1-tan^2x=\pm2\sqrt{3}tanx\Rightarrow\ tan^2x\pm2\sqrt{3}tanx-1=0$

and you've got to solve "both" quadratics.