Verified answer

x² - 2x sec α + tg² α = 0

x²-2.x/cos α + sen²α/cos²α=0

cos²α.x²-2.cos α.x + 1 - cos²α= 0

(cos α.x - 1)² = cos²α

cos α.x - 1 = cos α

cos α.(x - 1) =1

x=1/cos α.x - 1

x=sec α -1

delta = 4sec²α -4tg²α

x' = ´[2secα + 2(sec²α - tg²α)^0,5]/2 = secα + (sec²α -tg²α)^0,5

x' = secα + [(1 - sen²α)^0,5 ] / cosα = secα +1

x'' = secα -1

conjunto solução:

CS = {secα+1 ; secα -1 }

Até

x² - 2x secα + tg²α = 0

delta = (-2secα)² - 4.1.tg²α

delta = 4sec²α - 4tg²α, mas sec²α = tg²α + 1

delta = 4(tg²α + 1) - 4tg²α

delta = 4, raiz de 4 = 2

x1 = (2secα + 2)/2 = secα + 1

x2 = (2secα - 2)/2 = secα - 1

S:{secα + 1, secα - 1}