Hoe los je 3cscA-2sinA-5 = 0 op?

Antwoord:

# A = KPI + (- 1) ^ k (pi / 6), Kinz #

Uitleg:

# 3cscA-2sinA-5 = 0 #

# RArr3 / sina-2sinA-5 = 0 #

# RArr3-2sin ^ 2A-5sinA = 0 #

# RArr2sin ^ 2A + 5sinAcolor (rood) (- 3) = 0 #

# RArr2sin ^ 2A + 6sinA-sina-3 = 0 #

# RArr2sinA (sina + 3) -1 (sina + 3) = 0 #

#rArr (sina + 3) (2sinA-1) = 0 #

# RArrsinA = -3! In -1,1, sina = 1 / 2in -1,1 #

# RArrsinA = sin (pi / 6) #

# RArrA = KPI + (- 1) ^ k (pi / 6), Kinz #

# RArrA = KPI + (- 1) ^ k (pi / 6), Kinz #

Antwoord:

# A = (npi) / 2 + -pi / 3, ninZZ #

#color (wit) (A) = n90 ^ circ + -60 ^ circ, ninZZ #

Uitleg:

Eerst vermenigvuldigen we alles met # Sina # sinds # CSCA = 1 / sina # en sinA / sinA = 1 #

#sinA (3cscA-2sinA-5) = sina (0) #

# 3-2sin ^ 2A-5sinA = 0 #

Plaatsvervanger # X = sina #

# ^ 2 + 2x 5x-3 = 0 #

#X = (- b + -sqrt (b ^ 2-4ac)) / 2 #

#color (wit) (x) = (- 5 + -sqrt (5 ^ 2-4 (2 * -3))) / 4 #

#color (wit) (x) = (- 5 + -sqrt (25 + 24)) / 4 #

#color (wit) (x) = (- 5 + -sqrt (49)) / 4 #

#color (wit) (x) = (- 5 + -7) / 4 #

#color (wit) (x) = (- 5-7) / 4 of (-5 + 7) / 4 #

#color (wit) (x) = - 12/4 of 2/4 #

#color (wit) (x) = - 3 of 1/2 #

Echter, # -1lesinAle1 # dus we moeten nemen #1/2#

# Sina = 1/2 #

# A = arcsin (1/2) = pi / 6- = 30 ^ circ, A = (5pi) / 6- = 150 ^ circ #

# A = (npi) / 2 + -pi / 3, ninZZ #

#color (wit) (A) = n90 ^ circ + -60 ^ circ, ninZZ #