What is the solution of the equation sin x + sin 2x = 1 + cos x - sin x ?

First, we'll shift all terms to one
side:

sin x + sin 2x - 1 - cos x + sin x =
0

We'll combine like
terms:

sin 2x + 2sin x - 1 - cos x =
0

We'll apply the double angle identity for the term sin
2x:

sin 2a = sin (a+a)=sina*cosa +
sina*cosa=2sina*cosa

Comparing, we'll
get:

sin 2x = 2sin x*cos
x

We'll re-write the
equation:

2sin x*cos x + 2sinx - cosx - 1 =
0

We'll factorize by 2sin x the first 2
terms:

2sinx(cos x + 1) - (cos x + 1) =
0

We'll factorize by (cos x +
1):

(cos x + 1)(2sin x - 1) =
0

We'll cancel each
factor:

cos x + 1 = 0

We'll
subtract 1 both sides:

cos x =
-1

x = +/-arccos (-1) + 2kpi

x
= +/-pi + 2kpi

2sin x - 1 =
0

We'll add 1 both sides:

2sin
x = 1

sin x = 1/2

x =
(-1)^k*arcsin (1/2) + kpi

x = (-1)^k*(pi/6) +
kpi

The solutions of the equation are given
by the reunion of sets: {+/-pi + 2kpi / k is integer}U{(-1)^k*(pi/6) + kpi/k is
integer}.