How do you solve 3 sec^2 x - 4 = 03sec2x4=0?

1 Answer
May 10, 2015

I would start by rewriting it in terms of the cosine function, since sec(x)=1/cos(x)sec(x)=1cos(x). Using this and doing a bit of rearranging makes the equation 3/cos^{2}(x)=43cos2(x)=4. Now divide both sides by 44 and multiply both sides by cos^{2}(x)cos2(x) to get cos^{2}(x)=3/4cos2(x)=34. This implies that cos(x)=\pm\sqrt{3}/2cos(x)=±32.

The solutions of the equation cos(x)=\sqrt{3}/2cos(x)=32 are x=pi/6+2n\pi=30^{\circ}+360^{\circ}nx=π6+2nπ=30+360n for n=0,\pm 1,\pm 2, \pm 3,\ldots and x=-pi/6+2n\pi=-30^{\circ}+360^{\circ}n for n=0,\pm 1,\pm 2, \pm 3,\ldots

The solutions of the equation cos(x)=-\sqrt{3}/2 are x=(5pi)/6+2n\pi=150^{\circ}+360^{\circ}n for n=0,\pm 1,\pm 2, \pm 3,\ldots and x=(7pi)/6+2n\pi=210^{\circ}+360^{\circ}n for n=0,\pm 1,\pm 2, \pm 3,\ldots