# trigonometry equation helpi'm not good in trig at all help! 3tan^2x=24(1-secx) x=?

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We have to solve 3tan^2x=24(1-sec x)

3*(tan x)^2 = 24* (1 - sec x)

use the fact that tan x = sin x / cos x and sec x = 1 / cos x

=> 3*(sin x)^2/(cos x)^2 = 24*(1 - 1/cos x)

=> (sin x)^2/(cos x)^2 = 8*(1 - 1/cos x)

=> (sin x)^2/(cos x)^2 = 8*(cos x - 1)/cos x

=> (sin x)^2/(cos x) = 8*(cos x - 1)

=> (1 - (cos x)^2) = 8*(cos x)(cos x - 1)

=> (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 = -1/9 and cos x = 1

=> x = arc cos(-1/9) and x = 0

=> x = 96.37 + n*360 and x = 0 + n*360

**The required values of x are x = 96.37 + n*360 and x = n*360**

We'll divide the equation by 3 both sides:

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}.