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

1 Answer
May 10, 2015

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

The solutions of the equation #cos(x)=\sqrt{3}/2# are #x=pi/6+2n\pi=30^{\circ}+360^{\circ}n# 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#