Asked by rajagopal
find the value of x if sin2x+3cosx=2 then cos3x+sec3x is
Answers
Answered by
Steve
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.
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.
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