simple trig equation

**magmagod** · April 6th 2010, 02:46 PM

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??

**Deadstar** · April 6th 2010, 04:28 PM

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