Question

Solve the equation and write general solution: 2 + √3 sec x - 4 cos x = 2√3.

Answer 1

Given that

2 + √3 sec x - 4 cos x = 2√3

2 + √3 / cos x - 4 cos x = 2√3

Multiplying both sides by cos x , we get

2 cos x + √3  - 4 cos² x = 2√3

0 = 4 cos² x - 2 cos x + 2√3 - √3

4 cos² x - 2 cos x + 2√3 - √3  = 0

4 cos² x - 2 cos x + √3 = 0

Using Quadratic Formula, we get

$cosx=\frac{1\pm\sqrt{1-4\sqrt{3} } }{4}$

On solving it, we get imaginary roots.

So there is no Real Solution for this equation.

Answer 2

Solution:
Note: There might some typing mistake in the equation because on solving this the following result appears.

To solve the equation
$2 + \sqrt{3} sec \: x - 4 \: cos \: x = 2 \sqrt{3}$
as we know that

$cos \: x = \frac{1}{sec \: x}$
put this value in the equation,so that entire equation convert into one trigonometric function

$2 + \sqrt{3} sec \: x - \frac{4}{sec \: x} = 2 \sqrt{3} \\ \\ 2sec \: x + \sqrt{3} {sec}^{2}x - 4 = 2 \sqrt{3} sec \: x \\ \\ \sqrt{3} {sec}^{2} x + 2(1 - \sqrt{3} )sec \: x - 4 = 0$
as it is a Quadratic equation in sec x so it can be solved by factorisation or by Quadratic formula

$sec \: x = \frac{ - 2(1 - \sqrt{3}) + - \sqrt{4 {(1 - \sqrt{3}) }^{2} + 16 \sqrt{3} } }{3} \\ \\ = \frac{ - 2 + 2 \sqrt{3} + - \sqrt{4(1 + 3 - 2 \sqrt{3}) + 16 \sqrt{3} } }{3} \\ \\ = \frac{ - 2 + 2 \sqrt{3} + - \sqrt{16+8 \sqrt{3} } }{3} \\ \\ = \frac{ - 2 + 2 \sqrt{3} + - 2\sqrt{4+2 \sqrt{3} } }{3}$
$sec \: x = \frac{ - 2 + 2 \sqrt{3} + 2 \sqrt{4 + 2 \sqrt{3} } }{3} \\ \\ sec \: x = \frac{ - 2 + 2 \sqrt{3} - 2 \sqrt{4 + 2 \sqrt{3} } }{3}$