# simple trig equation

• Apr 6th 2010, 02:46 PM
magmagod
simple trig equation
ok so i have been doing these trig equations for a while now...and i came up with this suspicious easy one..can anybody tell me if what im doing is right?

cos^2x = 1

took the square root of both sides and ended up with cos x = 1... then i took the inverse of cos (1) and came up with 0.. now im checking when cos is positive, and it turns out its in Q 1 and 4

so in Q1 its the same as the related acute which is 0 and in Q 4 its 2 pi - 0 which gives 2 pi...

is this right? is it ok to take the square root of both sides to remove the squared in the cosine??
• Apr 6th 2010, 04:28 PM
Quote:

Originally Posted by magmagod
ok so i have been doing these trig equations for a while now...and i came up with this suspicious easy one..can anybody tell me if what im doing is right?

cos^2x = 1

took the square root of both sides and ended up with cos x = 1... then i took the inverse of cos (1) and came up with 0.. now im checking when cos is positive, and it turns out its in Q 1 and 4

so in Q1 its the same as the related acute which is 0 and in Q 4 its 2 pi - 0 which gives 2 pi...

is this right? is it ok to take the square root of both sides to remove the squared in the cosine??

You missed some stuff...

if $\displaystyle \cos^2(x) = 1$, $\displaystyle \cos(x) = \pm 1$.

Your answers of $\displaystyle 0$ and $\displaystyle 2 \pi$ were correct, however, it will also be equal to $\displaystyle 1$ at $\displaystyle \pm 2\pi$, $\displaystyle \pm 4\pi$, etc...

So $\displaystyle \cos(x) = 1$ when $\displaystyle x = \pm 2n\pi$, $\displaystyle n = 0,1,2,3, \dots$.

But we also have solutions when $\displaystyle \cos(x) = -1$, this happens at $\displaystyle \pm \pi$, $\displaystyle \pm \frac{3\pi}{2}$, $\displaystyle \pm \frac{5\pi}{2}$, etc...

I.e. when $\displaystyle x = \pm \frac{(2n+1)\pi}{2}$, $\displaystyle n=0,1,2,3,\dots$.
• Apr 6th 2010, 04:32 PM
Here's a graph of $\displaystyle \cos^2(x)$. Not.e where the graph is equal to 1. i.e. at the top of the 'waves'.