2sin(𝑥)cos(𝑥)+√3sin(𝑥)=0

Question

Find all angles in radians that satisfy the equation. For each solution enter first the angle solution in [0,2𝜋) then the period. When 2 or more solutions are available enter them in increasing order of the angles. (e.g. 𝑥=𝜋/2+2𝑘𝜋 or 𝑥=3𝜋/2+2𝑘𝜋 etc.)

Answer format is in x=_____+____ and there are four different answers which is the most confusing part to me.

Answer 1

2sinxcosx + sqr3sinx =0

factor out sinx

sinx(2cosx +sqr3) = 0

set each factor = 0

sinx=0, x=0 or pi

2cosx +sqr3 = 0

cosx =-(sqr3)/2

x = 5pi/6 or 7pi/6

x= 0, 5pi/6, pi, or 7pi/6

2pi is not a solution because of the closed parenthesis instead of a bracket for the interval [0, 2pi) Had it been [0, 2pi] then 2pi would be a 5th solution. For the same reason, there's no +2kpi where k = any integer, as that adds in more solutions that are outside the given interval. In degrees the solutions are 0, 150, 180 and 210.