Question 445499
{{{(sinx)2-3(cosx)^2=0}}}, substitute {{{(sinx)^2=1-(cosx)^2}}}, and get:

{{{1-(cosx)^2-3(cosx)^2=0}}} => {{{4(cosx)^2=1}}} => {{{(cosx)^2=1/4}}}. 

The last equation is equivalent with two equations:

{{{cosx=1/2}}} and {{{cosx=-1/2}}}. The solution for the first equation is:

{{{x=pi/3}}} and {{{x=5pi/3}}}. The solution for the second equation is:

{{{x=2pi/3}}} and {{{x=4pi/3}}}.