Cosx + SiNx = sqrt (cosx)?

Cevap:

# Rarrx = 2npi # nerede # ZZ'de #

Açıklama:

# Rarrcosx + sinx = sqrtcosx #

# Rarrcosx-sqrtcosx = -sinx #

#rarr (cosx-sqrtcosx) ^ 2 = (- SiNx) ^ 2 #

# Rarrcos ^ 2x-2cosx * sqrtcosx + cosx = sin ^ 2x = 1-cos ^ 2x #

# Rarr2cos ^ 2x-2cosx * sqrtcosx + cosx-1 = 0 #

let # Sqrtcosx = y # sonra # Cosx = y ^ 2 #

# Rarr2 * (y ^ 2) ^ 2-2 * y ^ 2 * y + y ^ 2-1 = 0 #

# Rarr2y ^ 4-2y ^ 3 + y ^ 2-1 = 0 #

# Rarr2y ^ 3 (y-1) + (y + 1) * (y-1) = 0 #

#rarr y-1 2y ^ 3 + y + 1 = 0 #

Alarak, # Rarry-1 = 0 #

# Rarrsqrtcosx = 1 #

# Rarrcosx = 1 = CoS0 #

# Rarrx = 2npi + -0 = 2npi # nerede # ZZ'de # genel olan

çözüm # X #.