Ako riešite 1 - 2 (sinx) ^ 2 = cosx, 0 <= x <= 360. Vyriešte x?

odpoveď:

# X = 0120240360 #

vysvetlenie:

# Asin ^ 2x + ACOS ^ 2x- = a #

# 1-2sin ^ 2x = 2cos ^ # 2x

# 1 (2-2cos ^ 2x) = cosx #

# 1-2 + 2cos ^ 2x = cosx #

# 2cos ^ 2x-cosx-1 = 0 #

náhradka # U = cosx #

# 2u ^ 2u-1 = 0 #

# U = (1 + -sqrt ((- 1) ^ 2-4 (2 * -1))) / (2 * 2) #

# U = (1 + -sqrt (1-4 (-2))) / 4 #

# U = (1 + -sqrt (1 + 8)) / 4 #

# U = (1 + -sqrt (9)) / 4 #

# U = (1 + -3) / 4 #

# U = 1- alebo-1/2 #

# Cosx = 1- alebo-1/2 #

# X = cos ^ -1 (1) = 0, (360-0) = 0360 #

# X = cos ^ -1 (-1/2) = 120, (360-120) = 120240 #

# X = 0120240360 #

odpoveď:

#color (modrá) (0, 120 ^ @, 240 ^ @, 360 ^ @) #

vysvetlenie:

identita:

#COLOR (červená) BB (sin ^ 2x + cos ^ 2x = 1) #

dosadením # (1-cos ^ 2x) # v danej rovnici:

# 1-2 (1-cos ^ 2 x) = cosx #

odčítanie # # Cosx a rozširovanie:

# 1-2 + 2cos ^ 2x-cosx = 0 #

zjednoduší:

# 2cos ^ 2x-cosx-1 = 0 #

nechať # = cosx #

#:.#

# 2u ^ 2u-1 = 0 #

Factor:

# (2u + 1) (u-1) = 0 => u = -1 / 2 a u = 1 #

ale # U = cosx #

#:.#

# cosx = -1 / 2, cosx = 1 #

# X = ARccOS (cosx) = ARccOS (-1/2) => x = 120 ^ @ #

Toto je v kvadrante II, máme tiež uhol v kvadrante III:

#360^@-120^@=240^@#

# x = arccos (cosx) = arccos (1) => x = 0, 360 ^ @ #

Riešenia zberu:

#color (modrá) (0, 120 ^ @, 240 ^ @, 360 ^ @) #