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View Full Version : Solve cos2x-cos^2x-2sinx+3=0 in the interval [0, 2 pi]


rocknnrebellion
Jan 21, 2010, 06:23 PM
solve cos2x-cos^2x-2sinx+3=0 in the interval [0, 2 pi]

ebaines
Jan 22, 2010, 06:35 AM
Hint: use a couple of the usual trig identities to get everything in terms of sin(x). For example:


cos(2x) = 1 - 2sin^2x \\cos^2x = 1 - sin^2x


Then you can solve for sin(x).