Another easy?trig question

1. Sep 29, 2009

zeion

1. The problem statement, all variables and given/known data

Solve the following equations of inequalities in the interval [0, 2pi)

cosx + 1 = sinx

2. Relevant equations

3. The attempt at a solution

cos2x + 2cosx + 1 = sin2x
cos2x + 2cosx + 1 = 1 - cos2x
2cos2x + 2cosx = 0
(2cosx)(cosx + 1) = 0
2cosx = 0 or cosx + 1 = 0
2cosx = 0
cosx = 0
x = pi/2, 3pi/2

cosx + 1 = 0
cosx = -1
x = pi

Did I miss anything?

2. Sep 30, 2009

mrkuo

You forgot to check your answers. When squaring an equation, you create the possibility of extraneous solutions appearing. In the case of your problem, 3pi/2 does not work in the original equation and should be left out of the solution set.

3. Sep 30, 2009

Chewy0087

yeh you can't be too careful squaring both sides as you risk getting extra solutions, perhaps a 'cleaner' way to do it is simply to rearrange giving;

sinx - cosx = 1

then proceed to right it in the form Rsin(A-B)= 1

but your method works, like mrko said just remember to check