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

תשובה:

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

# rarrx = (4n + 1) pi / 2 # איפה # nrarrZ #

הסבר:

#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 (x + 75 ^ @) - sin15 ^ @ = 0 #

# # rarr2sin (x + 75 ^ @ - 15 ^ @ @) / 2) cos (x + 75 ^ @ + 15 ^ @) / 2) = 0 #

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

או #rarrsin ((x + 60 ^ @) / 2) = 0 #

#rarr (x + 60 ^ @) / 2 = npi #

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

או, #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 #

תשובה:

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

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

הסבר:

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

#andcostheta! = 0 #, חלוקת שני הצדדים על ידי # costheta #

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

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

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

מוסיף # (I) ו- (II) #,אנחנו מקבלים.# 2sectheta = 4 => sectheta = 2 #

#color (אדום) (costheta = 1/2> 0) # #, מן equn נתון.

# costheta = 1/2 => (2 + sqrt3) (1/2) = 1-sintheta ## => 1 + sqrt (3) / 2 = 1-sintheta => צבע (אדום) (sintheta = -sqrt (3) / 2 <0) #

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