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How to solve the inequality sin2x>sin4x, if 0=<x=<2pi?
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First, we'll have to move sin 4x to the left side:
sin 2x - sin 4x > 0
The difference of the matching trigonometric functions returns the product:
sin 2x - sin 4x = 2 sin[(2x-4x)/2]*cos[(2x+4x)/2]
sin 2x - sin 4x = 2 sin(-x)*cos(3x)
Since the sine function is odd, the function sin(-x) = - sin x
sin 2x - sin 4x = 0
We'll re-write the equation:
-2 sin(x)*cos(3x) = 0
We'll cancel each factor:
sin x = 0
x = (-1)^k*arcsin 0 + k*pi
x = k*pi
cos(3x) = 0
3x = +/-arccos 0 + 2k*pi
3x = +/- (pi/2) + 2k*pi
x = +/- (pi/6) + 2k*pi/3
Now, we'll see where the functions sin x and cos 3x are both positive, or both negative, for the product to be strictly positive.
The functions sine and cosine are both positive in the 1st quadrant, and they are both negative in the 3rd quadrant.
sin x = 0 => x = 0 (1st quadrant)
sin x = 0 => x = pi (3rd quadrant)
cos 3x = 0 => 3x = pi/2 => x = pi/6 (1st quadrant)
cos 3x = 0 => 3x = pi + pi/2 => x = pi/3 + pi/6
x = 3pi/6 => x = 3pi/2 (3rd quadrant)
For the inequality to hold, the values of x belong to the reunion of intervals (0;pi/6)U(pi/2;5pi/6)U(3pi/2;11pi/6)
Posted by giorgiana1976 on July 1, 2011 at 3:24 PM (Answer #1)
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