Kako rešiti sin (2x) cos (x) = sin (x)?

Kako rešiti sin (2x) cos (x) = sin (x)?
Anonim

Odgovor:

# x = npi, 2 npi + - (pi / 4) in 2npi + - ((3pi) / 4) # kje #n v ZZ #

Pojasnilo:

# rarrsin2xcosx = sinx #

# rarr2sinx * cos ^ 2x-sinx = 0 #

#rarrsinx (2cos ^ 2x-1) = 0 #

# rarrrarrsinx * (sqrt2cosx + 1) * (sqrt2cosx-1) = 0 #

Kdaj # sinx = 0 #

# rarrx = npi #

Kdaj # sqrt2cosx + 1 = 0 #

# rarrcosx = -1 / sqrt2 = cos ((3pi) / 4) #

# rarrx = 2npi + - ((3pi) / 4) #

Kdaj # sqrt2cosx-1 = 0 #

# rarrcosx = 1 / sqrt2 = cos (pi / 4) #

# rarrx = 2npi + - (pi / 4) #

Odgovor:

#x = npi, pi / 4 + npi, (3pi) / 4 + npi # kje #n v ZZ #

Pojasnilo:

Imamo, #barva (bela) (xxx) sin2xcosx = sinx #

#rArr 2sinxcosx xx cosx = sinx # Kot, #sin 2x = 2sinxcosx #

#rArr 2sinxcos ^ 2x - sin x = 0 #

#rArr sinx (2cos ^ 2 - 1) = 0 #

Zdaj, Ali, #sin x = 0 rArr x = sin ^ -1 (0) = npi #, kje #n v ZZ #

Ali, #barva (bela) (xxx) 2cos ^ 2x - 1 = 0 #

#rArr 2cos ^ 2x - (sin ^ 2x + cos ^ 2x) = 0 # Kot # sin ^ 2x + cos ^ 2 x = 1 #

#rArr 2cos ^ 2x-sin ^ 2x-cos ^ 2x = 0 #

#rArr cos ^ 2x - sin ^ 2x = 0 #

#rArr (cosx + sin x) (cos x - sin x) = 0 #

Torej, Eno #cos x - sin x = 0 rArr cos x = sin x rArr x = pi / 4 + - npi #, kje #n v ZZ #

Ali, #cos x + sin x = 0 rArr cos x = -sinx rArr x = (3pi) / 4 + - npi #, kje #n v ZZ #

Torej, vse skupaj, #x = npi, pi / 4 + - npi, (3pi) / 4 + - npi #, kje #n v ZZ #