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}
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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}