find the value of x if sin2x+3cosx=2 then cos3x+sec3x is

Answer 1

sin2x+3cosx = 2
2sinxcosx+3cosx = 2
cosx(2sinx+3) = 2
2sinx+3 = secx

cos3x = cos2xcosx-sin2xsinx
= cos2xcosx-(2-3cosx)(2-3cosx)/(2cosx)
= (2cos^2x-1)cosx-(2-3cosx)^2/(2cosx)

I don't see where this is going. I fear that I am missing something here if you expect some simple answer.