Hanapin ang eksaktong halaga? 2sinxcosx + sinx-2cosx = 1

Hanapin ang eksaktong halaga? 2sinxcosx + sinx-2cosx = 1
Anonim

Sagot:

# rarrx = 2npi + - (2pi) / 3 # O # x = npi + (- 1) ^ n (pi / 2) # kung saan # nrarrZ #

Paliwanag:

# rarr2sinx * cosx + sinx-2cosx = 1 #

#rarrsinx (2cosx + 1) -2cosx-1 = #

#rarrsinx (2cosx + 1) -1 (2cosx + 1) = 0 #

#rarr (2cosx + 1) (sinx-1) = 0 #

Alinman, # 2cosx + 1 = 0 #

# rarrcosx = -1 / 2 = -cos (pi / 3) = cos (pi- (2pi) / 3) = cos ((2pi) / 3) #

# rarrx = 2npi + - (2pi) / 3 # kung saan # nrarrZ #

O, # sinx-1 = 0 #

# rarrsinx = 1 = sin (pi / 2) #

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