Paano mo malutas ang 2cos2x-3sinx = 1?

Sagot:

# x = arcsin (1/4) + 360 ^ circ k or #

# x = (180 ^ circ - arcsin (1/4)) 360 ^ circ k or #

#x = -90 ^ circ + 360 ^ circ k # para sa integer # k #.

Paliwanag:

# 2 cos 2x - 3 sin x = 1 #

Ang kapaki-pakinabang na double angle formula para sa cosine dito ay

#cos 2x = 1 - 2 sin ^ 2 x #

# 2 (1 - 2 sin ^ 2 x) - 3 sin x = 1 #

# 0 = 4 sin ^ 2 x + 3 sin x - 1 #

# 0 = (4 sin x - 1) (sin x 1) #

# sin x = 1/4 o sin x = -1 #

# x = arcsin (1/4) + 360 ^ circ k or x = (180 ^ circ - arcsin (1/4)) 360 ^ circ k or x = para sa integer # k #.

Sagot:

# rarrx = npi + (- 1) ^ n * sin ^ (- 1) (1/4) o npi + (- 1) ^ n * (- pi / 2) # # nrarrZ #

Paliwanag:

# rarr2cos2x-3sinx-1 = 0 #

# rarr2 (1-2sin ^ 2x) -3sinx-1 = 0 #

# rarr2-4sin ^ 2x-3sinx-1 = 0 #

# rarr4sin ^ 2x + 3sinx-1 = 0 #

#rarr (2sinx) ^ 2 + 2 * (2sinx) * (3/4) + (3/4) ^ 2- (3/4) ^ 2-1 = 0 #

#rarr (2sinx + 3/4) ^ 2 = 1 + 9/16 = 25/16 #

# rarr2sinx + 3/4 = + - sqrt (25/16) = + - (5) / 4 #

# rarr2sinx = + - 5 / 4-3 / 4 = (+ - 5-3) / 4 #

#rarrsinx = (+ - 5-3) / 8 #

Pagkuha # + ve # sign, makuha namin

# rarrsinx = (5-3) / 8 = 1/4 #

# rarrx = npi + (- 1) ^ n * sin ^ (- 1) (1/4) # # nrarrZ #

Pagkuha # -ve # sign, makuha namin

#rarrsinx = (- 5-3) / 8 = -1 #

# rarrx = npi + (- 1) ^ n * (- pi / 2) # kung saan # nrarrZ #