Riješite (2 + sqrt3) cos theta = 1-sin theta?

Odgovor:

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

# Rarrx = (4n + 1) pi / 2 # Gdje # NrarrZ #

Obrazloženje:

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

# Rarrtan75 ^ '* cosx + sinx = 1 #

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

# Rarrsinx * cos75 ^ + cosx * sin75 ^ '^' = cos75 = sin (90 '^ - ^ 15') = sin15 ^ '#

#rarrsin (x + 75 ^ ') - sin15 ^' = 0 #

# Rarr2sin ((x + 75 ^ '- ^ 15') / 2) cos ((x + 75 + 15 ^ '^') / 2) = 0 #

#rarrsin ((x + 60 ^ @) / 2) + cos ((x + 90 ^ ') / 2) = 0 #

Ili #rarrsin ((x + 60 ^ ') / 2) = 0 #

#rarr (x + 60 ^ ') / 2 = NPI #

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

ili, #cos ((x + 90 ^ ') / 2) = 0 #

#rarr (x + 90 ^ ') / 2 = (2n + 1) pi / 2 #

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

Odgovor:

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

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

Obrazloženje:

# (2 + sqrt3) costheta = 1 sintheta #

#andcostheta! = 0 #, dijeleći obje strane # Costheta #

# 2 + sqrt3 = sectheta-tantheta => sectheta-tantheta = 2 + sqrt3 do (I) #

#:. 1 / (sectheta-tantheta) = 1 / (2 + sqrt3) ## => (Sec ^ 2 theta-tan ^ 2 theta) / (sectheta-tantheta) = 1 / (2 + sqrt3) + (2-sqrt3) / (2-sqrt3) #

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

Dodavanje # (I) i (II) #, dobivamo.# 2sectheta = 4 => sectheta = 2 #

#COLOR (crveno) (costheta = 1/2> 0) #, Od danog equna.

# Costheta = 1/2 => (2 + sqrt3) (1/2) = 1 sintheta ## => 1 + sqrt (3) / 2 = 1 sintheta => boja (crvena) (= sintheta -sqrt (3) / 2 <0) #

# Theta = 2kpi-pi / 3 kinZ #,………. # (IV ^ (TH) #kvadrant)