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

Question

Find all the angles in radians that satisfy the equation.  For each solution enter first the angle solution in [0, 2pi), then the period.  When 2 or more solutions are available enter them in increasing order.

Answer 1

Factor out sin(x)

to get:

sin(x) [2cos(x)+1]=0

set each factor = 0

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

solve for x in each equation

sinx=0

inverse sin(0) = the angle whose sine is zero = 0 or pi

2cosx +1 = 0

2cosx = -1

cosx = -1/2

inverse cos(-1/2) = the angle whose cosine is -1/2 = 5pi/6 or 7pi/6

There are 4 solutions: x = 0, pi, 5pi/6 and 7pi/6

but in increasing order they are 0, 5pi/6, pi, 7pi/6

2pi could have been a solution but it does not fall within [0,2pi) given the parenthesis after 2pi