Question

How do you solve `sinx + cosx = 1` over the interval 0 to 2pi?

Answer 1

0, `pi/2` and `2pi`..

Explanation:

With t = tan `(x/2)`, use sin x = `(2t)/(1+t^2) and cos x = (1-t^2)/(1+t^2)`.
Then, t`(t-1)` = 0. t = tan `(x/2)` = 0 gives x = 0, 2`pi`.
t = tan`(x/2)` = 1 gives x = `pi/2`..