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

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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!

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