find-solution-8-tan-x-8tan-5-x-sec-6-x-in-x-0-pi-2-

Question Number 83975 by jagoll last updated on 08/Mar/20

find solution

8 \tan x - 8 \tan^{5} x = \sec^{6} x in x \in (0, \frac{π}{2})

Answered by TANMAY PANACEA last updated on 08/Mar/20

8 a - 8 a^{5} = {(1 + a^{2})}^{3}

8 a (1 - a^{4}) - {(1 + a^{2})}^{3} = 0

8 a (1 + a^{2}) (1 + a) (1 - a) - {(1 + a^{2})}^{3} = 0

(1 + a^{2}) {8 a (1 - a^{2}) - 1 - 2 a^{2} - a^{4}} = 0

1 + a^{2} \neq 0

8 a - 8 a^{3} - 1 - 2 a^{2} - a^{4} = 0

a^{4} + 8 a^{3} + 2 a^{2} - 8 a + 1 = 0

d e v i d i n g b y a^{2}

(a^{2} + \frac{1}{a^{2}}) + 8 (a - \frac{1}{a}) + 2 = 0

{(a - \frac{1}{a})}^{2} + 2 + 8 (a - \frac{1}{a}) + 2 = 0

k^{2} + 8 k + 4 = 0 [k = a - \frac{1}{a}]

k = \frac{- 8 \pm \sqrt{64 - 16}}{2} = \frac{- 8 \pm 4 \sqrt{3}}{2} = (- 4 \pm 2 \sqrt{3})

a - \frac{1}{a} = k

a^{2} - a k - 1 = 0

a = \frac{k \pm \sqrt{k^{2} + 4}}{2}

w a i t \dots

Answered by MJS last updated on 08/Mar/20

8 \tan x - {8 \tan}^{5} x = \sec^{6} x

8 \frac{\sin x}{\cos x} - 8 \frac{\sin^{5} x}{\cos^{5} x} = \frac{1}{\cos^{6} x} \Rightarrow \cos x \neq 0

8 \sin x \cos^{5} x - {8 \sin}^{5} x \cos x = 1

2 \sin 4 x = 1

\Rightarrow x = \frac{π}{24} \lor x = \frac{5 π}{24}

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