Question 1159951


{{{cos(x) cos (6x) =-1}}}

-> you know that {{{-1<=cos(x)<=1}}}


case 1: when {{{cos(x)=-1}}} and {{{cos(6x) =1}}}


{{{cos(x)=-1}}} 
{{{x=cos^-1(-1)}}}
{{{x=2pi*n+pi}}}

and {{{cos (6x) =1}}}
general solutions for {{{cos (6x )=1}}} is: 
{{{6x=2pi*n}}}
{{{x=(2pi*n)/6 }}}
{{{x=(pi*n)/3}}}


check: if {{{n=1}}}
{{{x=2pi+pi=3pi}}}... for {{{cos(x)}}}
{{{x=pi/3}}} ... for {{{cos (6x )}}}
{{{cos(3pi)*cos(6(pi/3)) =-1}}}->{{{true}}}


case 2: when {{{cos(x)=1}}} and {{{cos(6x) =-1}}}

{{{cos(x)=1}}}
{{{x=cos^-1(1)}}}
{{{x=2pi*n}}}

{{{cos(6x) =-1}}}
{{{6x=2pi*n+pi}}}
{{{x=2pi*n/6 +pi/6}}}
{{{x=(pi*n)/3+pi/6}}}


check: if {{{n=1}}}
{{{x=2pi}}} ...for {{{cos(x)}}}
{{{x=pi/3+pi/6=pi/2}}}... for {{{cos(6x)}}}

{{{cos(2pi)*cos(6(pi/2)) =-1}}}->{{{true}}}