Math, asked by gurugowthamibyreddy, 1 year ago

√cos2x+√1+sin2x=2√sinx+cosx if​

Answers

Answered by amitnrw
3

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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