Memecahkan 1 / (tan2x-tanx) -1 / (cot2x-cotx) = 1?

# 1 / (tan2x-tanx) -1 / (cot2x-cotx) = 1 #

# => 1 / (tan2x-tanx) -1 / (1 / (tan2x) -1 / tanx) = 1 #

# => 1 / (tan2x-tanx) + 1 / (1 / (tanx) -1 / (tan2x)) = 1 #

# => 1 / (tan2x-tanx) + (tanxtan2x) / (tan2x-tanx) = 1 #

# => (1 + tanxtan2x) / (tan2x-tanx) = 1 #

# => 1 / tan (2x-x) = 1 #

# => tan (x) = 1 = tan (pi / 4) #

# => x = npi + pi / 4 #

Menjawab:

# x = npi + pi / 4 #

Penjelasan:

# tan2x-tanx = (sin2x) / (cos2x) -sinx / cosx = (sin2xcosx-cos2xsinx) / (cos2xcosx) #

= #sin (2x-x) / (cos2xcosx) = sinx / (cos2xcosx) #

dan # cot2x-cotx = (cos2x) / (sin2x) -cosx / sinx = (sinxcos2x-cosxsin2x) / (sin2xsinx) #

= #sin (x-2x) / (sin2xsinx) = - sinx / (sin2xsinx) #

Karenanya # 1 / (tan2x-tanx) -1 / (cot2x-cotx) = 1 # dapat ditulis sebagai

# (cos2xcosx) / sinx + (sin2xsinx) / sinx = 1 #

atau # (cos2xcosx + sin2xsinx) / sinx = 1 #

atau #cos (2x-x) / sinx = 1 #

atau # cosx / sinx = 1 # yaitu # cotx = 1 = cot (pi / 4) #

Karenanya # x = npi + pi / 4 #