Find the solutions of this equation on the interval [0, 2π): sin(3π/2 + x) + sin(3π/2 + x) = -2

Question

please show your work and explain it simply.

Answer 1

sin(270+x) = -1

270+x = sin^-1(1)= the angle whose sine = -1 = 270 degrees

x = 270-270 = 0 degrees = -0 radians + 2npi where n=any integer =2npi

but you want an answer in the interval [0,2pi)

so that's x=0 radians or 0 degrees

2pi would also have been a solution, but the interval [0,2pi) does not include 2pi due to the lack of a bracket

Answer 2

If the problem was copied correctly,

sin(3π/2+x) + sin(3π/2+x) = 2sin(3π/2+x) =-2

Divide both sides by 2

sin(3π/2+x) = -1

On the interval [0,2π], sin(θ)=-1 only when θ=3π/2, or θ = -π/2

So,

3π/2+x = 3π/2 --> x = 0

3π/2+x = -π/2 -->x = -2π which is the same as 2π