Cosx + sinx = sqrt (cosx)?

대답:

# rarrx = 2npi # 어디에 ZZ # #n

설명:

# 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 #

방해 # sqrtcosx = y # 그때 # 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 #

복용, # rarry-1 = 0 #

# rarrsqrtcosx = 1 #

# rarrcosx = 1 = cos0 #

# rarrx = 2npi + -0 = 2npi # 어디에 ZZ # #n 일반

해결책 #엑스#.