How do you solve #sin4x-2sin2x=0# in the interval [0,360]?

Question

4942 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-09-19T01:44:15

Start by writing #sin4x# as #sin(2x + 2x)# and expand using the sum formula.

#sin(2x + 2x) - 2sin2x = 0#

The sum formula is #sin(A + B) = sinAcosB + cosAsinB#:

#sin2xcos2x + cos2xsin2x - 2sin2x = 0#

#2sin2xcos2x - 2sin2x = 0#

#2sin2x(cos2x - 1) = 0#

#sin2x = 0" AND "cos2x = 1#

#2x = 0˚, 180˚, 360˚, 540˚, 720˚" AND "2x = 0˚, 360˚, 720˚#

#x = 0˚, 90˚, 180˚, 270˚, 360˚#

Hopefully this helps!