How do you solve (sec(x) - 2)*(sqrt(3)*sec(x) - 2) = 0?

Question

in the interval [0,2pi)

Answer 1

You have the product of two binomials equal to zero. This means either binomial can be zero to make the equation true. So we have:
sec(x) - 2 = 0 sqrt(3)*sec(x) - 2 = 0.

The first part:
sec(x) - 2 = 0,
sec(x) = 2,
1/cos(x) = 2,
cos(x) = 1/2.
Do we know any value of 'x' for which this equation holds? I hope so: cos(pi/3) = 1/2 and cos(5*pi/3) = 1/2.

The second part (very similar):
sqrt(3)*sec(x) - 2 = 0,
sqrt(3)*sec(x) = 2,
sec(x) = 2/sqrt(3),
1/cos(x) = 2/sqrt(3),
cos(x) = sqrt(3)/2.
Do your little circle-diagram-thingy to see that:
cos(pi/6) = sqrt(3)/2 and cos(11*pi/6) = sqrt(3)/2.

So your solution set is:
{pi/3, 5*pi/3, pi/6, 11*pi/6}

Feel free to email me for any clarifications.