√cos2x+√1+sin2x=2√sinx+cosx if
Answers
Given : √cos2x+√1+sin2x=2√sinx+cosx
To find : x x ∈ [ 0 , 2 π ]
Solution:
√cos2x+√(1+sin2x)=2√(sinx+cosx)
using cos2x = cos²x - Sin²x
Cos²x + Sin²x = 1
sin2x = 2SinxCosx
=> √(cos²x - Sin²x) + √( Cos²x + Sin²x + 2SinxCosx) = 2 √(sinx+cosx)
=> √((cosx + Sin x)(cosx - Sin x)) + √( Cosx + Sinx)² = 2 √(sinx+cosx)
=> √( Cosx + Sinx) ( √(cosx - Sin x) + √( Cosx + Sinx) ) - 2 √(sinx+cosx) = 0
taking common √( Cosx + Sinx)
=> √( Cosx + Sinx) (√( Cosx- Sinx) + √( Cosx + Sinx) - 2) = 0
=> √( Cosx + Sinx) = 0 or √( Cosx- Sinx) + √( Cosx + Sinx) - 2 = 0
=> Cosx + Sinx =0
=> Cosx = - sinx => tanx = - 1 => x = 3π/4 , 5π/4
√( Cosx - Sinx) + √( Cosx + Sinx) - 2 = 0
=> √( Cosx - Sinx) + √( Cosx + Sinx) = 2
Squaring both sides
cosx - Sinx + cosx + sinx + 2 √( Cos²x - Sin²x) = 4
=> 2Cosx + 2 √Cos2x = 4
=> Cosx + √Cos2x = 2
=> x = 0 or 2π
Hence x = 0 , 3π/4 , 5π/4 , 2π
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