Solve {2 + 2sin2x} / {2 (1 + sinx) (1-sinx)} = sec ^ 2x + tanx?

Sagot:

#x = k pi quad # integer # k #

Paliwanag:

Lutasin # {2 + 2sin2x} / {2 (1 + sinx) (1-sinx)} = sec ^ 2x + tanx #

# 0 = {2 + 2sin2x} / {2 (1 + sinx) (1-sinx)} - sec ^ 2x - tanx #

# = {2 + 2 (2 sin x cos x)} / {2 (1-sin ^ 2 x)} - 1 / cos ^ 2x - sin x / cos x #

# = {1 + 2 sinx cos x} / {cos ^ 2 x} - 1 / cos ^ 2 x - {sin x cos x} / cos ^ 2 x #

# = {sin x cos x} / {cos ^ 2 x} = tan x #

# tan x = 0 #

#x = k pi quad # integer # k #

Sagot:

# x = kpi, kinZZ #

Paliwanag:

Meron kami, # (2 + 2sin2x) / (2 (1 + sinx) (1-sinx)) = sec ^ 2x + tanx #

# => (2 (1 + sin2x)) / (2 (1-sin ^ 2x)) = sec ^ 2x + tanx #

# => (1 + sin2x) / cos ^ 2x = sec ^ 2x + tanx #

# => 1 + sin2x = sec ^ 2xcos ^ 2x + tanxcos ^ 2x #

# => 1 + sin2x = 1 + sinx / cosx xxcos ^ 2x #

# => sin2x = sinxcosx #

# => 2sin2x = 2sinxcosx #

# => 2sin2x = sin2x #

# => 2sin2x-sin2x = 0 #

# => kulay (pula) (sin2x = 0 … hanggang (A) #

# => 2x = kpi, kinZZ #

# => x = (kpi) / 2, kinZZ #

Ngunit, para dito # x #,# sinx = 1 => 1-sinx = 0 #

Kaya, # (2 + 2sin2x) / (2 (1 + sinx) (1-sinx)) = (2 + 0) / (2 (1 + 1) (0)) = 2 / hindi natukoy

Kaya,

#x! = (kpi) / 2, kinZZ #

Kaya, walang solusyon. !!

Muli mula sa # (A) #

# sin2x = 0 => 2sinxcosx = 0 => sinxcosx = 0 #

# => sinx = 0 o cosx = 0, kung saan, tanx at secx # ay tinukoy.

# i.e. cosx! = 0 => sinx = 0 => color (violet) (x = kpi, kinZZ #

May pagkakasalungatan sa resulta kapag nakuha natin # sin2x = 0 #.