Circ functions help

[lem]

a. Show that sin 2x + sin4x = sin2x(3-4sin^2x)
b. Hence solve sin2x + sin4x=0   x  [0,2pie]

[brightsky]

a. Show that sin 2x + sin4x = sin2x(3-4sin^2x)

sin2x + sin4x
= sin2x + 2sin2xcos2x
= sin2x(1 + 2cos2x)
= sin2x(1 + 2(1-2sin^2(x)))
= sin2x(1 + 2 - 4 sin^2(x))
= sin2x (3 - 4 sin^2(x))
as req.

b. Hence solve sin2x + sin4x=0   x  [0,2pie]

sin2x (3-4sin^2(x)) =0
sin2x = 0 or sin^2(x) = 3/4
solve as per usual.

[lem]

Thank you