Cos3x+sin5x=0.............

Question

Cos3x+sin5x=0.............

Answer 1

Sin (pi/2 - 3x) + sin 5x = 0
2 sin[(pi/2 + 2x)/2]cos[(pi/2 - 8x)/2] = 0
sin[(pi/2 + 2x)/2] = 0
pi/4 + x = k*pi
x = -pi/4 + k*pi
cos[(pi/2 - 8x)/2] = 0
pi/4 - 4x = pi/2 + k*pi
4x = -pi/4 - k*pi
x = -pi/16 - k*pi/4