Determine the general solution, in radians, for sin(2θ)sinθ+cos^2 θ=1 by solving algebraically

Answer 1

sin2x(sinx) + cos^2(x) = 1

2sinxcosxsinx + cos^2(x) = cos^2(x)+ sin^2(x)

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

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

sin^2(x) = 0

sinx = 0, x = 0, pi

cosx = 1/2, x=pi/3, 5pi/3

x= 0, pi/3,5pi.3

Answer 2

1=cos²θ+sin²θ

sin2θ=2cosθsinθ

have

2cosθsin²θ-sin²θ=0

so either sinθ=0

or cosθ=1/2

θ=0,π/3,π,5π/6