How to solve tanx + cotx = 4cos2x? Thank you

Question

1457 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-02-18T11:01:12

General solution: #x=(4n+1)pi/8 ; n in z#

#tanx+cotx=4cos2x or sinx/cosx+cosx/sinx=4cos2x # or

#(sin^2x+cos^2x)/(cosx*sinx)=4cos2x # or

#1/(1/2(2cosx*sinx))=4cos2x ; [sin2x=2sinxcosx] # or

#2/(sin2x)=4cos2x or 2sin2xcos2x=1#

#[sin4x=2sin2xcos2x]# or

#sin4x=1; sin (pi/2)=1 :. 4x =pi/2 :. x=pi/8#

General solution: #x=(4n+1)pi/8 ; n in z# [Ans]