Question

Solve the following equations for all reals.

sin^2 x - sin x = 0
sin 2x + cos x = 0

Answer 1

For the first equation, factor out a sinx.
sinx(sinx-1)=0
Set both factors equal to 0
So, find when sinx=0 and sinx=1
so x= 0, π, 2π, and π/2.
Add a +2πn to each answer since there's an infinite number if it's asking for all reals.

x= 0+2πn, π+2πn, π/2 + 2πn

Answer 2

factor out a sin x huh? I dont know what you mean sinx * sinx -sinx =0?

Answer 3

For the second equation, you have to know that sin2x = 2sinxcosx, so you have 2sinxcosx + cosx =0

Factor out a cosx.
cosx(2sinx + 1) = 0
Set both factors equal to 0, so
cosx=0 and sinx= -1/2
x= π/2 + 2πn, 3π/2 + 2πn
x= 7π/6 + 2πn, 11π/6 + 2πn

Answer 4

For the first equation, you have sin^2x, or sinx squared, - sinx. You need to factor out a sinx and you get sinx*(sinx-1)=0 (you can make sure the equations are equivalent by foiling.

You also need to factor for the second equation.