Question

Find cos(x) if sin^2 x-1/cos x = -1

Answer 1

do you know that \[1-\sin^2(x)=\cos^2(x)\]?

Answer 2

no I am not sure

Answer 3

oh, well it does

Answer 4

Oh I thought I had to solve for something. That is just a trig function sorry lol

Answer 5

the mother of all trig identities is \[\cos^2(x)+\sin^2(x)=1\] and all its cousins
for example if you subtract \(\sin^2(x)\) you get \[1-\sin^2(x)=\cos^2(x)\]

Answer 6

you don't have \[1-\sin^2(x)\]you have \[\sin^2(x)-1\] which is the same as \[-\cos^2(x)\]

Answer 7

So you basically just have to simplify the trig function but not to a number?

Answer 8

you still have to solve the equation for \(x\)

Answer 9

but since the numerator is \(-\cos^2(x)\) it should be relatively easy

Answer 10

How do I solve for x? I have an idea but it's not like a normal equation

Answer 11

put \(-\cos^2(x)\) in the top, then cancel a cosine

Answer 12

on the top of what? like a fraction?

Answer 13

or set it equal to -1?

Answer 14

@jim_thompson5910

Answer 15

The original equation is this

\[\Large \frac{\sin^2(x)-1}{\cos(x)} = -1\]

right? or am I off?

Answer 16

OR is it this

\[\Large \sin^2(x)-\frac{1}{\cos(x)} = -1\]

@sammietaygreen

Answer 17

thanks

Answer 18

so as @satellite73 is saying, you'd follow these steps

\[\Large \frac{\sin^2(x)-1}{\cos(x)} = -1\]

\[\Large \frac{-\cos^2(x)}{\cos(x)} = -1\]

\[\Large -\cos(x) = -1\]

\[\Large \cos(x) = 1\]

Answer 19

so do I just solve x when it is simplified to cos(x) = 1

Answer 20

no need to solve for x itself

they just want the value of `cos(x)`

Answer 21

Ohhh I get it now I don't know why I didn't understand before. So you just had to simplify it to get 1

Answer 22

yep 1 goes in the box

Answer 23

Okay. I just didn't understand what he was understanding about x but I get what you're saying. Thank you very much.