Hoe los je sin (x) - cos (x) -tan (x) = -1 op?

Hoe los je sin (x) - cos (x) -tan (x) = -1 op?
Anonim

Antwoord:

# "De oplossingsset" = {2kpi} uu {kpi + pi / 4}, k in ZZ #.

Uitleg:

Gezien dat, # SiNx-cosx-tanx = -1 #.

#:. SiNx-cosx-sinx / cosx + 1 = 0 #.

#:. (SiNx-cosx) - (sinx / cosx-1) = 0 #.

#:. (SiNx-cosx) - (sinx-cosx) / cosx = 0 #.

#:. (SiNx-cosx) cosx- (SiNx-cosx) = 0 #.

#:. (SiNx-cosx) (cosx-1) = 0 #.

#:. sinx = cosx of cosx = 1 #.

# "Casus 1:" sinx = cosx #.

Observeer dat #cosx! = 0, want "anders wordt" tanx "wordt" #

undefined.

Vandaar delen door #cosx! = 0, sinx / cosx = 1, of, tanx = 1 #.

#:. tanx = tan (pi / 4) #.

#:. x = kpi + pi / 4, k in ZZ, "in dit geval" #.

# "Geval 2:" cosx = 1 #.

# "In dit geval" cosx = 1 = cos0,:. x = 2kpi + -0, k in ZZ #.

Alles bij elkaar hebben we

# "De oplossingsset" = {2kpi} uu {kpi + pi / 4}, k in ZZ #.

Antwoord:

# Rarrx = 2npi, NPI + pi / 4 # waar #n in ZZ #

Uitleg:

# Rarrsinx-cosx-tanx = -1 #

# Rarrsinx-cosx-sinx / cosx + 1 = 0 #

#rarr (SiNx * cosx-cos ^ 2x aSiNx + cosx) / cosx = 0 #

# Rarrsinx * cosx aSiNx-cos ^ 2x + cosx = 0 #

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

#rarr (cosx-1) (SiNx-cosx) = 0 #

Wanneer # Rarrcosx-1 = 0 #

# Rarrcosx = cos0 #

# Rarrx = 2npi + -0 = 2npi # waar #n in ZZ #

Wanneer # Rarrsinx-cosx = 0 #

#rarrcos (90-x) -cosx = 0 #

# Rarr2sin ((90-x + x) / 2) * sin ((x-x + 90) / 2) = 0 #

#rarrsin (x-pi / 4) = 0 # Zoals #sin (pi / 4)! = 0 #

# Rarrx-pi / 4 = npi #

# Rarrx = npi + pi / 4 # waar #n in ZZ #