Question 436796
{{{6 cos3(x) + 6 cos(x) = 0}}} ==>  cos3x +  cosx = 0

==> cos2x cosx - sin2x sinx + cosx = 0
==> {{{cos2x cosx - 2(sinx)^2* cosx  + cosx = 0}}}
==> {{{cosx(cos2x - 2(sinx)^2  + 1) = 0}}}
==> {{{cosx(1 - 2(sinx)^2 - 2(sinx)^2 + 1) = 0}}}
==> {{{cosx(2 - 4(sinx)^2) = 0}}}
==> {{{cosx(1-2(sinx)^2) = 0}}}
==> cosx = 0 ==> x = {{{pi/2}}}, {{{(3pi)/2}}};
==> Also, {{{(sinx)^2 = 1/2}}}, or {{{sinx = sqrt(2)/2}}}, {{{sinx = -sqrt(2)/2}}}
==> x = {{{pi/4}}}, {{{(3pi)/4}}}, {{{(5pi)/4}}}, {{{(7pi)/4}}}.

==> Solution set is { {{{pi/2}}}, {{{(3pi)/2}}}, {{{pi/4}}}, {{{(3pi)/4}}}, {{{(5pi)/4}}}, {{{(7pi)/4}}} }.