Επίλυση (2 + sqrt3) cos theta = 1-sin theta;

Απάντηση:

# rarrx = (6n-1) * (pi / 3) #

# rarrx = (4η + 1) πΙ / 2 # Οπου # nrarrZ #

Εξήγηση:

# rarr (2 + sqrt (3)) cosx = 1-sinx #

# rarrtan75 ^ @ * cosx + sinx = 1 #

# rarr (sin75 ^ @ cosx) / (cos75 ^ @) + sinx = 1 #

# rarrsinx * cos75 ^ @ + cosx * sin75 ^ @ = cos75 ^ @ = sin (90 ^ @ - 15 ^ @) = sin15 ^

# rarrsin (χ + 75 ^ @) - sin15 ^ @ = 0 #

# rarr2sin ((χ + 75 ^ - 15 ^) / 2) cos ((χ + 75 ^ + 15 ^) /

# rarrsin ((χ + 60 ^ @) / 2) * cos ((χ + 90 ^ @) / 2) = 0 #

Είτε # rarrsin ((χ + 60 ^ @) / 2) = 0 #

# rarr (χ + 60 ^ @) / 2 = npi #

# rarrx = 2npi-60 ^ = 2npi-pi / 3 = (6n-1) * (pi / 3)

ή, #cos ((x + 90 ^ @) / 2) = 0 #

# rarr (χ + 90 ^ @) / 2 = (2n + 1) pi / 2 #

# rarrx = 2 * (2n + 1) pi / 2-pi / 2 = (4n + 1) pi /

Απάντηση:

Αν, # costheta = 0 => sintheta = 1 => theta = (4k + 1) pi / 2, kinZ #

# theta = 2kpi-pi / 3, kinZ #,

Εξήγηση:

# (2 + sqrt3) costheta = 1-sintheta #

#andcostheta! = 0 #, διαιρώντας και τις δύο πλευρές # costheta #

# 2 + sqrt3 = sectheta-tantheta => sectheta-tantheta = 2 + sqrt3 έως (Ι) #

#:. 1 / (sectheta-tantheta) = 1 / (2 + sqrt3) #(2-sqrt3) / (2-sqrt3) # (2-sqrt3) / (2-sqrt3)

# => sectheta + tantheta = 2-sqrt3 έως (II) #

Προσθέτωντας # (Ι) και (II) #,παίρνουμε.# 2sectheta = 4 => sectheta = 2 #

#color (κόκκινο) (costheta = 1/2> 0) #, Από το δεδομένο equn.

# costheta = 1/2 => (2 + sqrt3) (1/2) = 1-sintheta ## = 1 + sqrt (3) / 2 = 1-sintheta => χρώμα (κόκκινο) (sintheta = -sqrt (3) / 2 <

# theta = 2kpi-pi / 3, kinZ #,………. # (IV ^ (th) #τεταρτοκύκλιο)