# evaluating solutions to trig equation??

• Aug 22nd 2008, 01:19 AM
Craka
evaluating solutions to trig equation??
$\displaystyle \begin{array}{l} {\rm Trying to find all solutions to interval inclusive of - pi to + pi for } \\ 2 + \cos (2x) = 3\cos (x) \\ {\rm What I have done is this:} \\ 2 + [\cos ^2 (x) - \sin ^2 (x)] = 3\cos (x) \\ 2 + \cos ^2 (x) - \sin ^2 (x) = 3\cos (x) \\ 2 + \cos ^2 (x) - 3\cos (x) = \sin ^2 (x) \\ \end{array}$

$\displaystyle \begin{array}{l} 2 + \cos ^2 (x) - 3\cos (x) = 1 - \cos ^2 (x) \\ 2 + 2\cos ^2 (x) - 3\cos (x) = 1 \\ 2\cos ^2 (x) - 3\cos (x) + 1 = 0 \\ (2\cos (x) - 2)(\cos (x) - 1) = 0 \\ {\rm So then }2\cos (x) = 2{\rm and }\cos (x) = 1 \\ {\rm Which gives one root of cos(x) = 1} \\ \end{array}$
• Aug 22nd 2008, 01:28 AM
Moo
Hello,
Quote:

Originally Posted by Craka
$\displaystyle \begin{array}{l} {\rm Trying to find all solutions to interval inclusive of - pi to + pi for } \\ 2 + \cos (2x) = 3\cos (x) \\ {\rm What I have done is this:} \\ 2 + [\cos ^2 (x) - \sin ^2 (x)] = 3\cos (x) \\ 2 + \cos ^2 (x) - \sin ^2 (x) = 3\cos (x) \\ 2 + \cos ^2 (x) - 3\cos (x) = \sin ^2 (x) \\ \end{array}$

$\displaystyle \begin{array}{l} 2 + \cos ^2 (x) - 3\cos (x) = 1 - \cos ^2 (x) \\ 2 + 2\cos ^2 (x) - 3\cos (x) = 1 \\ 2\cos ^2 (x) - 3\cos (x) + 1 = 0 \\ (2\cos (x) - 2)(\cos (x) - 1) = 0 \\ {\rm So then }2\cos (x) = 2{\rm and }\cos (x) = 1 \\ {\rm Which gives one root of cos(x) = 1} \\ \end{array}$

Something went wrong in your factorisation :)

$\displaystyle (2 \cos(x)-2)(\cos(x)-1)=2(\cos(x)-1)^2=2 \cos^2(x)-{\color{red}4}\cos(x)+{\color{red}2}$

$\displaystyle 2 \cos^2(x)-3\cos(x)+1=(2 \cos(x)-{\color{magenta}1})(\cos(x)-1)=0$

This gives $\displaystyle \cos(x)=\frac 12$ or $\displaystyle \cos(x)=1$.
• Aug 22nd 2008, 07:41 AM
Soroban
Hello, Craka!

There are three forms for $\displaystyle \cos2\theta$

. . $\displaystyle \cos2\theta \;=\;\begin{Bmatrix}\cos^2\!\theta - \sin^2\!\theta \\ 2\cos^2\!\theta - 1 \\ 1 - 2\sin^2\!\theta \end{Bmatrix}$

The last two can be derived from the first,
but I've found it expedient to memorize all three.

Quote:

Find all solutions on $\displaystyle [-\pi,\:+\pi]\!:\;\;2 + \cos (2x) \:=\:3\cos (x)$

We have: . $\displaystyle 2 + \cos2x \:=\:3\cos x$

. . . . . $\displaystyle 2 + \overbrace{2\cos^2\!x - 1} \:=\:3\cos x$

. . $\displaystyle 2\cos^2\!x - 3\cos x + 1 \:=\:0$

.$\displaystyle (\cos x - 1)(2\cos x - 1) \:=\:0$

Therefore: . $\displaystyle \begin{array}{ccccc}\cos x - 1 \:=\:0 & \Rightarrow & \cos x \:=\:1 & \Rightarrow &\boxed{x \:=\:0} \\ \\[-3mm] 2\cos x - 1 \:=\:0 & \Rightarrow & \cos x \:=\:\frac{1}{2} & \Rightarrow & \boxed{x \:=\:\pm\frac{\pi}{3}} \end{array}$