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sin3θ = cos2θ find the most general values of θ satisfying the equatios?
sinax + cosbx = 0 solve ?
Dear Mayank,
sin (3x) = sin(2x + x) = sin(2x)cos(x) + cos(2x)sin(x) = 2sin(x)cos(x)cos(x) + (1 - 2sin² (x) ) sin(x) = 2sin(x)cos² (x) + sin(x) - 2sin³ (x) = 2sin(x) ( 1 -sin² (x) ) + sin(x) - 2sin³ (x) = 2sin(x) -2sin³ (x) + sin(x) - 2sin³ (x) = -4sin³ (x) + 3sin(x) = 3sin(x) - 4sin³ (x)
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