Trigonometric Equations Question?

Question

Solve the following equation on the interval [0, 2Pi). Do not use a calculator.

sinx + 2sin (x/2) = cos (x/2) + 1

Answer 1

sinx + 2sin (x/2) = cos (x/2) + 1

sin x = 2 sin(x/2) cos(x/2)
2 sin(x/2) cos(x/2) + 2sin (x/2) - cos (x/2) - 1=0
2 sin(x/2) {cos(x/2) +1} -{ cos (x/2) + 1} =0
{ cos (x/2) + 1} {2sin(x/2) -1 } =0

Either { cos (x/2) + 1} =0 or
{2sin(x/2) -1 } =0

a) cos (x/2) + 1 ---> cos(x/2) = -1 ---> x/2 = Pi --> x = 2 Pi (rejected because 2Pi is not in the domain)

b) 2sin(x/2) -1 =0 --> sin(x/2) = 1/2
x/2 = Pi/6 or 5Pi/6

x = Pi/3 or x = 5Pi/3

Answer 2

let t=x/2
the equation becomes
sin2t+2sint=cost + 1
2sintcost+2sint=cos t +1
2sint(cost+1)=cost+1
divide by cos t+1 both sides
2sint=1
sint=1/2
t = pi/6 or 5pi/6
but t=x/2 => x=2t
x=pi/3 or 5pi/3

Answer 3

Id look up half angle trig identities