Kako najti splošno rešitev 5 sin (x) +2 cos (x) = 3?

Kako najti splošno rešitev 5 sin (x) +2 cos (x) = 3?
Anonim

Odgovor:

# rarrx = npi + (- 1) ^ n * (sin ^ (- 1) (3 / sqrt29)) - sin ^ (- 1) (2 / sqrt29) # #n inZZ #

Pojasnilo:

# rarr5sinx + 2cosx = 3 #

#rarr (5sinx + 2cosx) / (sqrt (5 ^ 2 + 2 ^ 2)) = 3 / (sqrt (5 ^ 2 + 2 ^ 2) #

# rarrsinx * (5 / sqrt (29)) + cosx * (2 / sqrt (29)) = 3 / sqrt29 #

Let # cosalpha = 5 / sqrt29 # potem # sinalpha = sqrt (1-cos ^ 2alpha) = sqrt (1- (5 / sqrt29) ^ 2) = 2 / sqrt29 #

Tudi, # alpha = cos ^ (- 1) (5 / sqrt29) = sin ^ (- 1) (2 / sqrt29) #

Zdaj se dani enačbi spremeni v

# rarrsinx * cosalpha + cosx * sinalpha = 3 / sqrt29 #

#rarrsin (x + alpha) = sin (sin ^ (- 1) (3 / sqrt29)) #

# rarrx + sin ^ (- 1) (2 / sqrt29) = npi + (- 1) ^ n * (sin ^ (- 1) (3 / sqrt29)) #

# rarrx = npi + (- 1) ^ n * (sin ^ (- 1) (3 / sqrt29)) - sin ^ (- 1) (2 / sqrt29) # #n inZZ #

Odgovor:

#x = 12 ^ @ 12 + k360 ^ @ #

#x = 124 ^ @ 28 + k360 ^ @ #

Pojasnilo:

5sin x + 2cos x = 3.

Obe strani delite s 5.

#sin x + 2/5 cos x = 3/5 = 0,6 # (1)

Pokličite #tan t = sin t / (cos t) = 2/5 # --> #t = 21 ^ @ 80 # -> cos t = 0,93.

Enačba (1) postane:

#sin x.cos t + sin t.cos x = 0.6 (0.93) #

#sin (x + t) = sin (x + 21,80) = 0,56 #

Kalkulator in krog enote podajata dve rešitvi za (x + t) ->

a. x + 21,80 = 33,92

#x = 33.92 - 21.80 = 12 ^ @ 12 #

b. x + 21,80 = 180 - 33,92 = 146,08

#x = 146.08 - 21.80 = 124 ^ @ 28 #

Splošni odgovori:

#x = 12 ^ @ 12 + k360 ^ @ #

#x = 124 ^ @ 28 + k360 ^ @ #

Preverite s kalkulatorjem.

#x = 12 ^ @ 12 # -> 5sin x = 1,05 -> 2cos x = 1,95

5sin x + 2cos x = 1.05 + 1.95 = 3. Dokazano.

#x = 124 ^ @ 28 # -> 5sin x = 4.13 -> 2cos x = -1.13

5sin x + 2cos x = 4.13 - 1.13 = 3. Dokazano.