Question 978216
I am assuming that the equation is {{{24 sin^2 (x) = 24 − 12 cos (x)}}} .

I would divide both sides of the equation by {{{12}}} to get a friendlier equivalent equation:
{{{24 sin^2 (x) = 24 -12 cos (x)}}}--->{{{24 sin^2 (x)/12 = (24 - 12 cos (x))/12}}}--->{{{2sin^2 (x) = 2 - cos (x)}}} .
Then, using the fact that
{{{sin^2(x)+cos^2(x)=1}}}<--->{{{sin^2(x)=1-cos^2(x)}}} ,
I would re-write {{{2sin^2 (x) = 2 - cos (x)}}} as
{{{2(1-cos^2(x)) = 2 - cos (x)}}} ,
and then I would just write {{{u}}} for {{{cos(x)}}} to save ink.
With the "change of variable" {{{u=cos(x)}}} , I save ink and confusion:
{{{2(1-u^2) = 2-u}}} looks a lot easier on the eyes.
Now, I solve:
{{{2(1-u^2) = 2 - u}}} 
{{{2-2u^2 = 2 - u}}}
{{{2-2u^2+2u^2-2 = 2 - u+2u^2-2}}}
{{{0=-u+2u^2}}}
That last equation would traditionally be written as
{{{2u^2-u=0}}} , and is a "quadratic equation".
Quadratic equations can be solved by "completing the square" or by using the quadratic formula.
In certain cases, they can also be solved by factoring,
and are that lucky in this case. Factoring we get
{{{2u^2-u=0}}}--->{{{u(2u-1)=0}}}--->{{{system(u=0,"or",2u-1=0)}}}--->{{{system(u=0,"or",u=1/2)}}} .
In the interval with {{{0<=x<2pi}}} ,
{{{u=cos(x)=0}}}--->{{{system(highlight(x=pi/2),"or",highlight(x=3pi/2))}}} and {{{u=cos(x)=1/2}}}--->{{{system(highlight(x=pi/3),"or",highlight(x=5pi/3))}}}