Wie lösen Sie sin (x) - cos (x) -tan (x) = -1?

Wie lösen Sie sin (x) - cos (x) -tan (x) = -1?
Anonim

Antworten:

# "The Solution Set" = {2kpi} uu {kpi + pi / 4}, k in ZZ #.

Erläuterung:

In Anbetracht dessen # 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 oder cosx = 1 #.

# "Fall 1:" sinx = cosx #.

Beobachte das #cosx! = 0, weil "wenn sonst;" tanx "wird" #

nicht definiert.

Daher teilen durch #cosx! = 0, sinx / cosx = 1 oder tanx = 1 #.

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

#:. x = kpi + pi / 4, k in ZZ, "in diesem Fall" #.

# "Fall 2:" cosx = 1 #.

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

Insgesamt haben wir, # "The Solution Set" = {2kpi} uu {kpi + pi / 4}, k in ZZ #.

Antworten:

# rarrx = 2npi, npi + pi / 4 # woher #n in ZZ #

Erläuterung:

# rarrsinx-cosx-tanx = -1 #

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

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

# rarrsinx * cosx-sinx-cos ^ 2x + cosx = 0 #

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

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

Wann # rarrcosx-1 = 0 #

# rarrcosx = cos0 #

# rarrx = 2npi + -0 = 2npi # woher #n in ZZ #

Wann # rarrsinx-cosx = 0 #

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

# rarr2sin ((90-x + x) / 2) * sin ((x-90 + x) / 2) = 0 #

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

# rarrx-pi / 4 = npi #

# rarrx = npi + pi / 4 # woher #n in ZZ #