Question

Please help? Precalc? Solve the equation for all real values of x. The equation is (1/(cosx-sinx))=cosx+sinx.

Answer 1

\[[answer: \pi k]\]
but what is the process for getting that?

Answer 2

\[\cos^2x-\sin^2x=1\]

Answer 3

an identity :
cos^2 (x) - sin^2 (x) = cos(2x)

Answer 4

Ohh so then it would be cos(2x)-1=0?

Answer 5

cos(2x)=1

Answer 6

yeah, 1 is cos(0)
so, cos(2x) = cos(0)
again, use the identity of
cos(x) = cos(x + k * 2pi)

Answer 7

cos(2x) = cos(0)
cos(2x) = cos(0 + k * 2pi)
2x = 0 + k * 2pi
2x = k * 2pi

x = k * pi

Answer 8

Ohh okay, that's what I couldn't figure out. I didn't know cos(0)=1. Thank you!

Answer 9

you're welcome