Paano mo malulutas ang kasalanan (2x) cos (x) = kasalanan (x)?

Paano mo malulutas ang kasalanan (2x) cos (x) = kasalanan (x)?
Anonim

Sagot:

# x = npi, 2npi + - (pi / 4), at 2npi + - ((3pi) / 4) # kung saan #n sa ZZ #

Paliwanag:

# rarrsin2xcosx = sinx #

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

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

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

Kailan # sinx = 0 #

# rarrx = npi #

Kailan # sqrt2cosx + 1 = 0 #

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

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

Kailan # sqrt2cosx-1 = 0 #

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

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

Sagot:

#x = npi, pi / 4 + npi, (3pi) / 4 + npi # kung saan #n sa ZZ #

Paliwanag:

Meron kami, #color (white) (xxx) sin2xcosx = sinx #

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

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

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

Ngayon, Alinman, #sin x = 0 rArr x = sin ^ -1 (0) = npi #, kung saan #n sa ZZ #

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

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

Kaya, Alinman #cos x - sin x = 0 rArr cos x = sin x rArr x = pi / 4 + - npi #, kung saan #n sa ZZ #

O kaya, #cos x + sin x = 0 rArr cos x = -sinx rArr x = (3pi) / 4 + - npi #, kung saan #n sa ZZ #

Kaya, Summing lahat ng ito, #x = npi, pi / 4 + - npi, (3pi) / 4 + - npi #, kung saan #n sa ZZ #