Find all solutions to the equation in the interval [0, 2π). sin 2x - sin 4x = 0?
QUESTION #2
Rewrite with only sin x and cos x.
cos 2x + sin x
THESE ARE THE CHOICES FOR QUESTION 2
1 + 3 sin x
1 + 3 sin2x
1 - 2 sin2x + sin x
1 + 2 sin2x + sin x
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sin(2x) - sin(4x) =>
sin(3x - x) - sin(3x + x) =>
sin(3x)cos(x) - sin(x)cos(3x) - sin(3x)cos(x) - sin(x)cos(3x) =>
-2sin(x)cos(3x)
-2 * sin(x) * cos(3x) = 0
sin(x) * cos(3x) = 0
sin(x) = 0
x = 0 , pi
cos(3x) = 0
3x = pi/2 + pi * k
3x = (pi/2) * (1 + 2k)
x = (pi/6) * (1 + 2k)
x = pi/6 , 3pi/6 , 5pi/6 , 7pi/6 , 9pi/6 , 11pi/6
x = pi/6 , pi/2 , 5pi/6 , 7pi/6 , 3pi/2 , 11pi/6
x = 0 , pi/6 , pi/2 , 5pi/6, pi , 7pi/6 , 3pi/2 , 11pi/6
cos(2x) + sin(x) =>
cos(x)^2 - sin(x)^2 + sin(x) =>
1 - 2sin(x)^2 + sin(x)