sin(2x)+cos(3x)=0

Question

solving between [0,2Π]

Answer 1

This problem can be solved in a much simpler way.

sin(-2x) = cos(3x)

=> cos(pi/2 + 2x) = cos(3x)

pi/2 + 2x = 3x => x = 0.5pi

or

pi/2 + 2x = -3x + 2k*pi

k = 1, x = 0.3pi

k = 2, x = 0.7pi

k = 3, x = 1.1pi

k = 4, x = 1.5pi

k = 5, x = 1.9pi

Answer 2

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