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

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We have to solve sin x + sin 2x = 1 + cos x - sin x

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

=> 2*sin x + sin 2x = 1 + cos x

=> 2*sin x + 2*sin x*cos = 1 + cos x

=> 2*sin x(1 + cos x) = 1 + cos x

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

2*sin x - 1 = 0

=> sin x = 1/2

x = 30 + n*360 degrees

cos x + 1 = 0

=> x = arc cos (-1)

=> x = 180 + n*360 degrees

**The solution of the equation is 30 + n*360 degrees and 180 + n*360 degrees.**

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}.**