Bagaimana anda menyelesaikan 1 - 2 (sinx) ^ 2 = cosx, 0 <= x <= 360. Menyelesaikan x?

Jawapan:

# x = 0,120,240,360 #

Penjelasan:

# 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 #

Pengganti # u = cosx #

# 2u ^ 2-u-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 = 1or-1/2 #

# cosx = 1or-1/2 #

# x = cos ^ -1 (1) = 0, (360-0) = 0,360 #

# x = cos ^ -1 (-1/2) = 120, (360-120) = 120,240 #

# x = 0,120,240,360 #

Jawapan:

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

Penjelasan:

Identiti:

#color (merah) bb (sin ^ 2x + cos ^ 2x = 1) #

Penggantian # (1-cos ^ 2x) # dalam persamaan yang diberi:

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

Mengurangkan # cosx # dan berkembang:

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

Mudahkan:

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

Biarkan # u = cosx #

#:.#

# 2u ^ 2-u-1 = 0 #

Faktor:

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

Tetapi # u = cosx #

#:.#

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

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

Ini dalam kuadran II, kami juga mempunyai sudut dalam kuadran III:

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

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

Mengumpul penyelesaian:

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