Find all solutions in radians

Question

a) 2sin^2x=sin x

b) 2cos^2x-cos x-1=0

c0tan^2x -tan x =0

Answer 1

a) 2sin^2x=sin x 2Sin^2x - sin x sinx(2Sinx -1) = 0 Sinx = 1/2 x = Π/6

b) 2cos^2x-cos x-1=0 (2cosx +1)(cosx -1) = 0 Cosx = -1/2 and cosx = 1, x = 4Π/3 and 0

c) tan^2x -tan x =0 tanx(tanx-1) = 0 x = 0 and 5Π/4

Answer 2

2sin^2(x) -sin(x) =0

sin(x)[2sin(x) -1] = 0

sin(x) = 0, x = 0 or pi

2sin(x)-1 =0

2sin(x) =1

sin(x) =1/2

x = 30 degrees or 150 degrees = pi/6 or 5pi/6

but the problems says find "all" solutions

so add in 360n to 30 & 150 and 2pin to pi/6 and 5pi/6 where n= any integer

in other words

x = 30, 390, -330, 750, ...etc

and x = 150, 410, 770, 1130, - 330, - 690 ... ect.

there are an infinite number of solutions

x=30 +360n, and 150 + 360n, where n = any integer, positive or negative

in radians

x =pi/6 + 2npi and 5pi/6 + 2npi where n= any integer