Find all solutions of cosx-1=sinx over the interval [0,2pi)

Answer 1

cosx - 1 = sinx

cosx = sinx + 1

sin^2x + cos^2x = 1

cos^2x = 1-sin^2x

cosx =

sqr(1-sin^2x) = sinx +1

square both sides (but you may introduce one extraneous solution not in the original equation)

1-sin^2 = sin^2x + 2sinx + 1

2sin^2x + 2sinx = 0

sinx(sinx +1) = 0

sinx = 0 or sinx=-1

x = 0 or pi, or 3pi/2

one may be a false extraneous solution

plug them into to the original equation to see if they work

cosx-1 = sinx

cos0 -1 = sin0

1-1 = 0

cospi-1 = sinpi

-1-1 doesn't = 0

pi is not a solution, just x=0

2pi is also not a solution since the interval has a parentheses not bracket for the interval upper bound

x=-3pi/2

cos(3pi/2) - 1 = sin(3pi/2)

0-1 = -1

There are two solutions, x=0 and x=3pi/2

or x= 0 and x = 270 degrees