sin2x+cosx=0 on the interval [0,2pi]?

Answer 1

Use the fact that sin(2x) = 2sin(x)cos(x), then factor

2sin(x)cos(x) + cos(x) = 0
(2sin(x) + 1) cos(x) = 0

So either 2sin(x)+1=0, or cos(x)=0. Take the two cases where sin(x) = -1/2 and the two cases where cos(x) = 0. This means x = π/2, 7π/6, 3π/2, 11π/6

Answer 2

sin2x = 2sinxcosx

2sinxcosx +cosx = 0

factor cosx out

cosx(2sinx +1) = 0

so, it is either:

cosx = 0 or 2sinx+1 = 0

if cosx = 0

then x = pi/2 and 3pi/2

if 2sinx +1 = 0

then sinx = -1/2

then x = 7pi/6 and 11pi/6


so your answers would be

x = pi/2, 3pi/2, 7pi/6 and 11pi/6


best answer?

Answer 3

=cosx(2sinx+1)=0
cosx=0 x=pi/2 and 3pi/2
sin x=-12
x=7pi/6 and 11pi/6