Question

@AZ #14

Answer 1

i factored it out to \[(2cosx+1)(cosx-1)\]
so cosx = -1/2 cosx=1
Which is 2pi/3 and also 0

Answer 2

so then that's \[\frac{ 2\pi }{ 3 } + 2\pi k\] and just 2pi k

Answer 3

How did you factor it to that?

Think of cos(x) as `a`

2a^2 - a = 1

how do you factor 'a' out ?

Answer 4

i actually put the 1 to the left, and then factored it

Answer 5

oh hmmm

Answer 6

Maybe that's how you're supposed to actually do it ;b

okay let me see

Answer 7

lolll, kk

Answer 8

so, yeah you were correct in factoring it. That was my mistake

So you should be getting multiple solutions

cos(x) = -1/2

you correctly got x = 2pi/3 + 2pi*k as a solution but there's one more
think about your unit circle, there's going to be two values where cos(x) can be -1/2

cos(x) = 1
and x = 2pi*k is correct for this

Answer 9

7pi/6

Answer 10

nope, that's when sin is -1/2

Answer 11

The unit circle lists (cos, sin)

Answer 12

4pi/3
oopssss

Answer 13

So that's the third solution + 2pi*k

Answer 14

thanksssssssss xdxd

Answer 15

you're welcome