1. ## 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??

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

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

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

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

I.e. when $x = \pm \frac{(2n+1)\pi}{2}$, $n=0,1,2,3,\dots$.

3. Here's a graph of $\cos^2(x)$. Not.e where the graph is equal to 1. i.e. at the top of the 'waves'.