Question

Guide me through this trig problem.

cos(x+pi/6)-1=cos(x-pi/6)

Answer 1

I'm assuming you solve for cos(x+pi/6) and cos(x-pi/6) first?

Answer 2

i'm getting

cos(x)(sqrt3/2)-sin(x)(1/2)-1=cos(x)(sqrt3/2)-sin(x)(1/2)

Answer 3

so is this just....-1?

Answer 4

@Michele_Laino @TheSmartOne

Answer 5

Check the right hand side again.
\[\cos(x-y)=\cos x\cos y\color{red}+\sin x\sin y\]

Answer 6

oh yeah, right

Answer 7

so we can cross out cos(x) and (sqrt3/2)?

Answer 8

Yes, your sine terms remain.

Answer 9

the (1/2) is crossed out as well?

Answer 10

the right side is:
cos(x)(sqrt3/2) + sin(x)(1/2)

Answer 11

No I meant the sine terms along with their coefficients. You have
\[-\frac{1}{2}\sin x-1=\frac{1}{2}\sin x\]
which you could divide through by \(\dfrac{1}{2}\) to get
\[-\sin x-2=\sin x\]

Answer 12

so sin(x)=-1 which would be 3pi/2?

Answer 13

That's one solution, yes. If you're restricted to \(0\le x<2\pi\), then you're done.

Answer 14

think i got it now, thank you.

Answer 15

yw