# 2tan²x + 2 sec²x - secx = 12

lemjay | Certified Educator

Express the equation as one trigonometric function.

`2tan^2x + 2sec^2x - secx =12`

From the trigonometric identity `1 + tan^2theta = sec^2theta` , replace `tan^2x` with `sec^2x-1` .

`2(sec^2x - 1) + 2sec^2x - secx = 12`

`2sec^2x - 2 + 2sec^2x - sec x = 12`

`4sec^2x -sec x - 2 = 12`

`4sec^2x - secx-14 = 0`

Then, express the equation in a form  `ax^2+bx + c = 0`  .  So,   let sec x = z.

`4z^2 - z - 14 =0`

Use the quadratic formula to solve for z.

`z=(-b+-sqrt(b^2-4ac))/(2a) = [-(-1)+-sqrt((-1)^2-4(4)(-14))]/(2(4))=(1+-sqrt225)/8 = (1+-15)/8`

The values of z are:

`z= (1+15)/8=16/8 = 2`

`z = (1-15)/8 = -14/8 = -7/4`

Substitute the values of z to sec x = z. Then, solve for x. Also, take note that sec x is the reciprocal of cos x.

`sec x = 2`

`1/(cosx) = 2`

`cos x = 1/2`

Note that cos x is positive at quadrants I and IV.

`x = 60 and 300` degrees

`sec x = -7/4`

`1/(cos x) =-7/4`

`cos x = -4/7`

Cosine function is negative at quadrants II and III.

`x = 124.85 and 235.15` degrees

Hence, the general solutions are:

`x_1 = 60 + 360k` degrees

`x_2 = 124.85 + 360k` degrees

`x_3 = 235.15 + 360k` degrees

`x_4 = 300 + 360k ` degrees

where k is an integer.