Question

(sin x) - 2 (sin x)^2 = 0. How do I solve the equation in the interval [0, 2pi) ?

Answer 1

factoring
sinx( 1 - 2sinx) = 0
sinx = 0 or 1- 2sinx = 0
sinx = 0 => x = 0, Pi, 2pi
1-2sinx = 0 => sinx = 1/2 => x = pi/6, 5pi/ 6

Answer 2

1) sinx=0... x=0,pi
2)sinx = 1/2 .. x= pi/6, 5pi/6
it can't be 2pi since the interval does not include 2pi

Answer 3

um yes, Nitin is right...I missed that part...so the answer included, 0, pi, pi/6 and 5pi/6

Answer 4

wait would it be 1/2 because you're dividing by the -2? or is santas_little_helper right?

Answer 5

1 - 2sinx = 0
1 = 2sinx => sinx = 1/2

Answer 6

okay i see now, thanks so much :)

Answer 7

:) no problem guys