does anyone know how to get this question? I got pi/2 and 3pi/3.
√2 sin² -sin = 0 Solve for values of sin. 0<x<2pi
Is it sin(x) for each?
sin(x) (√2sin²x - 1) = 0
sin(x) = 0 or sin²x = √2/2
x = 0, pi for the first, and sinx = √√2 / √2 which is 2^(3/4) / 2
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