Question

Find all solutions in the interval [0, 2pi) show all algebra 2sin^2 x-sinx= 1

Answer 1

2sin^2 x - sinx = 1
2sin^2 x - sinx - 1 = 0
(2sinx+1)(sinx-1) = 0
for the zeroes, take each factor then setting be zero
2sinx + 1 = 0 or sinx-1=0

for the 2sinx+1=0
2sinx = -1
sinx=-1/2
x = 210 , 330

for the sinx-1=0
sinx = 1
x = 90

sso, the solution are 90,210,330