solve the equation

cos3X sin3(cube)X + sin3X cos3(cube)X = 0

Dear somesh,

we know that,

cos^3x=(cos3x+3cosx)/4 and  sin^3x=(3sinx-sin3x)/4                    [since cos3x=4cos^x-3cosx]

substuting in the given equation

cos3X sin3(cube)X + sin3X cos3(cube)X  =  0

cos3x{(3sinx-sin3x)/4}+sin3x{(cos3x+3cosx)/4} = 0

{3sinxcos3x-cos3xsin3x+sin3xcos3x+3cosxsin3x}/4 = 0

3/4(sinxcos3x+cosxsin3x) = 0

sin4x = 0

Hence x=n∏/4

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