Find general solution of
sin^3x - cos^3x = 1 + sinx cosx
Urgent
sin^3x - cos^3x = (sinx - cosx)(sin^2x + sinx cosx + cos^2x) = 1 + sinx cosx
so the equation is really just
(1 + sinx cosx)(sinx - cosx - 1) = 0
Now you have two cases:
1 + 1/2 sin2x = 0
sin2x = -2
no joy there.
or,
sinx - cosx = 1
note that this is just
√2 (sinx * 1/√2 - 1/√2 cosx) = √2 sin(x - π/4)
so now we have
√2 sin(x - π/4) = 1
sin(x - π/4) = 1/√2
x - π/4 = π/4 or 3π/4
x = (π/2 or π) + 2kπ
or
sinx - cosx = 1
square both sides:
sin^2 x - 2sinxcosx + cos^2 x = 1
sin 2x = 0
2x = 0 or 2x = π or 2x = 2π or ....
x = 0 or x = π/2 or x = π
BUT, since we squared the equation , each solution must be verified
in sinx - cosx = 1
if x = 0,
LS = sin 0 - cos 0 = 0 - 1 = -1
RS = 1 + (sin0)(1) = 1 + 0(1) = 1 , so x ≠ 0
if x = π/2
LS = sin π/2 - cosπ/2 = 1 - 0 = 1
RS = 1 + (1)(0) = 1 , So x = π/2
if x = π
LS = 0 - (-1) = 1
RS = 1 + (0)(-1) = 0 , x = π
from there we get the same answer as oobleck
looks longer because of the checking, but does not depend
on a not-well-known formula