Menu Close

sin-2-4x-cos-2-x-2sin-4x-cos-2-x-




Question Number 135867 by liberty last updated on 16/Mar/21
sin^2 (4x)+cos^2 (x)=2sin (4x)cos^2 (x)
sin2(4x)+cos2(x)=2sin(4x)cos2(x)
Answered by EDWIN88 last updated on 16/Mar/21
recall cos^2 x = (1/2)+(1/2)cos 2x  ⇔ [ 2sin 2x cos 2x ]^2 +(1/2)+(1/2)cos 2x=4sin 2x cos 2x ((1/2)+(1/2)cos 2x)  4sin^2 2x cos^2 2x +(1/2)cos 2x+(1/2)=2sin 2x cos 2x+2sin 2x cos^2 2x  8sin^2 2x cos^2 2x + cos 2x + 1 = 4sin 2x cos 2x +4sin 2x cos^2 2x  8sin^2 2x cos^2 2x+cos 2x+sin^2 2x+cos^2 2x=4sin 2x cos 2x +4sin 2x cos^2 2x
recallcos2x=12+12cos2x[2sin2xcos2x]2+12+12cos2x=4sin2xcos2x(12+12cos2x)4sin22xcos22x+12cos2x+12=2sin2xcos2x+2sin2xcos22x8sin22xcos22x+cos2x+1=4sin2xcos2x+4sin2xcos22x8sin22xcos22x+cos2x+sin22x+cos22x=4sin2xcos2x+4sin2xcos22x
Answered by liberty last updated on 17/Mar/21
Answered by MJS_new last updated on 17/Mar/21
using trigonometric formulas I get  [sin x =s∧cos x =c]  −c^2 (64c^2 s^4 −16s^2 −1)=8c^3 s(2c^2 −1)  ⇒ cos x =0 ⇔ ★ x=nπ−(π/2)∧n∈Z ★  −(64c^2 s^4 −16s^2 −1)=8cs(2c^2 −1)  this has no real solution
usingtrigonometricformulasIget[sinx=scosx=c]c2(64c2s416s21)=8c3s(2c21)cosx=0x=nππ2nZ(64c2s416s21)=8cs(2c21)thishasnorealsolution

Leave a Reply

Your email address will not be published. Required fields are marked *