Solve the equation tan^2x=8-8secx.
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We have to solve (tan x)^2 = 8 - 8*sec x.
(tan x)^2 = 8 - 8*sec x
=> (sin x)^2 / (cos x)^2 = 8 - 8/(cos x)
=> (1 - (cos x)^2) / (cos x)^2 = (8*(cos x)^2 - 8* cos x)/ (cos x)^2
=> (1 - (cos x)^2) = (8*(cos x)^2 - 8* cos x)
=> 9(cos x)^2 - 8 cos x - 1 = 0
=> 9(cos x)^2 - 9 cos x + cos x - 1 = 0
=> 9(cos x)(cos x - 1) + 1( cos x - 1) = 0
=> (9(cos x) - 1)(cos x - 1) = 0
cos x = 1 and cos x = 1/9
x = arc cos 1 and x = arc cos (1/9) and - arc cos (1/9)
The required values are : x = 2*n*pi and x = arc cos (1/9) + 2*n*pi and x = -arc cos (1/9) + 2*n*pi
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We'll substitute (tan x)^2 = (sec x)^2 - 1
We'll re-write the equation:
(sec x)^2 - 1 = 8 - 8secx
We'll move all terms to one side:
(sec x)^2 - 1 - 8 + 8secx = 0
We'll put sec x = t
t^2 + 8t - 9 = 0
We'll write -9 = -1-8
t^2 + 8t - 8 - 1 = 0
We'll group the 1st and the last terms and the middle terms:
(t^2 - 1) + 8(t-1) = 0
We'll re-write the difference of squares:
(t-1)(t+1) + 8(t-1) = 0
(t-1)(t+1+8) = 0
(t-1)(t+9) = 0
We'll put t-1 = 0
t = 1
But t = sec x = 1/cos x
1/cos x = 1
cos x = 1
x = 2k*pi
t+9 = 0
sec x = -9
1/cos x = -9
cos x = -1/9
x = +/-arccos(1/9) + 2kpi, k is integer
The solutions of the equation are:{2k*pi} ; {+/-arccos(1/9) + 2kpi}.
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