Com es resol el sin (2x) cos (x) = sin (x)?

Resposta:

# x = npi, 2npi + - (pi / 4) i 2npi + - ((3pi) / 4) # on #n en ZZ #

Explicació:

# rarrsin2xcosx = sinx #

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

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

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

Quan # sinx = 0 #

# rarrx = npi #

Quan # sqrt2cosx + 1 = 0 #

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

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

Quan # sqrt2cosx-1 = 0 #

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

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

Resposta:

#x = npi, pi / 4 + npi, (3pi) / 4 + npi # on #n en ZZ #

Explicació:

Tenim, #color (blanc) (xxx) sin2xcosx = sinx #

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

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

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

Ara, Qualsevol, #sin x = 0 rArr x = sin ^ -1 (0) = npi #, on? #n en ZZ #

O, #color (blanc) (xxx) 2cos ^ 2x - 1 = 0 #

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

Així, bé #cos x - sin x = 0 rArr cos x = sin x rArr x = pi / 4 + - npi #, on? #n en ZZ #

O, #cos x + sin x = 0 rArr cos x = -sinx rArr x = (3pi) / 4 + - npi #, on? #n en ZZ #

Per tant, sumant-ho tot, #x = npi, pi / 4 + - npi, (3pi) / 4 + - npi #, on? #n en ZZ #