Paano mo malutas ang 3cscA-2sinA-5 = 0?

Sagot:

# A = kpi + (- 1) ^ k (pi / 6), kinZ #

Paliwanag:

# 3cscA-2sinA-5 = 0 #

# rArr3 / sinA-2sinA-5 = 0 #

# rArr3-2sin ^ 2A-5sinA = 0 #

# rArr2sin ^ 2A + 5sinAcolor (pula) (- 3) = 0 #

# rArr2sin ^ 2A + 6sinA-sinA-3 = 0 #

# rArr2sinA (sinA + 3) -1 (sinA + 3) = 0 #

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

# rArrsinA = -3! sa -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 #

Sagot:

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

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

Paliwanag:

Una naming paramihin ang lahat ng bagay # sinA # dahil # cscA = 1 / sinA # at sinA / sinA = 1 #

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

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

Kapalit # x = sinA #

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

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

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

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

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

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

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

#color (white) (x) = - 12/4 o 2/4 #

#color (white) (x) = - 3 o 1/2 #

Gayunpaman, # -1lesinAle1 # kaya kailangan namin #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 (white) (A) = n90 ^ circ + -60 ^ circ, ninZZ #