Vprašanje # bfc9a

Odgovor:

# x = 0,2pi #

Pojasnilo:

Vaše vprašanje je

#cos (x-pi / 6) + cos (x + pi / 6) = sqrt3 # v intervalu # 0,2pi #.

To vemo iz trigonomskih identitet

#cos (A + B) = cosAcosB-sinAsinB #

#cos (A-B) = cosAcosB + sinAsinB #

tako da daje

#cos (x-pi / 6) = cosxcos (pi / 6) + sinxsin (pi / 6) #

#cos (x + pi / 6) = cosxcos (pi / 6) -sinxsin (pi / 6) #

zato, #cos (x-pi / 6) + cos (x + pi / 6) #

# = cosxcos (pi / 6) + sinxsin (pi / 6) + cosxcos (pi / 6) -sinxsin (pi / 6) #

# = 2cosxcos (pi / 6) #

Zdaj vemo, da lahko poenostavimo enačbo

# 2cosxcos (pi / 6) = sqrt3 #

#cos (pi / 6) = sqrt3 / 2 #

tako

# sqrt3cosx = sqrt3 -> cosx = 1 #

To vemo v intervalu # 0,2pi #, # cosx = 1 # kdaj # x = 0, 2pi #

Odgovor:

# "No soln. In" (0,2pi) #.

Pojasnilo:

#cos (x-pi / 6) + cos (x + pi / 6) = sqrt3 #

Uporaba, # cosC + cosD = 2cos ((C + D) / 2) cos ((C-D) / 2) #, # 2cosxcos (-pi / 6) = sqrt3 #, #:. 2 * sqrt3 / 2 * cosx = sqrt3 #, #:. cosx = 1 = cos0 #.

Zdaj, # cosx = udobno rArr x = 2kpi + -y, k v ZZ #.

#:. cosx = cos0 rArr x = 2kpi, k v ZZ, tj.

# x = 0, + - 2pi, + -4pi, … #

": Soln. Set" pod (0,2pi) "je" phi #.