# evaluating solutions to trig equation??

• August 22nd 2008, 02:19 AM
Craka
evaluating solutions to trig equation??
$
\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}$

$
\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}$
• August 22nd 2008, 02:28 AM
Moo
Hello,
Quote:

Originally Posted by Craka
$
\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}$

$
\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 :)

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

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

This gives $\cos(x)=\frac 12$ or $\cos(x)=1$.
• August 22nd 2008, 08:41 AM
Soroban
Hello, Craka!

There are three forms for $\cos2\theta$

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

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

Quote:

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

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

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

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

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

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