# Trig Equations with Multiple Trig Functions cont.

• Apr 6th 2008, 05:23 PM
~berserk
Trig Equations with Multiple Trig Functions cont.
Find, to the nearest tenth of a degree, all values of x in the interval 0≤x≤360 that satisfy the equation.

tan x=cos x

also solve for x in the interval 0≤x≤2π

cos ½x= cos x
• Apr 7th 2008, 01:39 AM
janvdl
Quote:

Originally Posted by ~berserk
tan x=cos x

$\displaystyle \frac{sin x}{cos x} = cos x$

$\displaystyle sin x = cos^2 x$

$\displaystyle sin x = 1 - sin^2 x$

$\displaystyle sin^2 x + sin x - 1 = 0$

Solve for x. (Let Sin x = k and solve using the quadratic formula.)

Quote:

Originally Posted by ~berserk
cos ½x= cos x

$\displaystyle cos x - cos \frac{x}{2} = 0$

This is true where x is 0 and there also seems to be a solution between 45 and 60 degrees, as well as between 15 and 30 degrees. I suggest you draw an accurate graph to find the points.
• Apr 7th 2008, 08:41 AM
Mathstud28
for the second one I suggest doing it this way
$\displaystyle cos\bigg(\frac{x}{2}\bigg)=cos(x)$...now using the identity for $\displaystyle cos\bigg(\frac{x}{2}\bigg)$ you obtain $\displaystyle \sqrt{\frac{1+cos(x)}{2}}=cos(x)$...now by squaring each side and simplifying you get $\displaystyle 2cos(x)^2-1-cos(x)=0$ Define u $\displaystyle sin(x)$ and solve by quadratic formula
• Apr 7th 2008, 05:43 PM
~berserk
i think i messed up the second problem yesterday because I did it and solved it like this
• Apr 7th 2008, 05:50 PM
Mathstud28
Quote:

Originally Posted by ~berserk
i think i messed up the second problem yesterday because I did it and solved it like this
http://i182.photobucket.com/albums/x...serk02/eqn.jpg

IDk...but here you go you factored it right so therefore $\displaystyle cos(x)=1$...so $\displaystyle x=cos^{-1}(1)=0$....$\displaystyle cos\bigg(\frac{1}{2{\cdot{0}/bigg)=cos(0)$...and doing the other one you get $\displaystyle x=\frac{2\pi}{3}$ which is a valid solution