Question 983506


1. {{{2cos^2(2x)}}} - {{{cos(2x)}}} = {{{0}}}.


Introduce new variable   {{{y}}} = {{{cos(2x)}}}. 


Then your equation will be 


{{{2y^2}}} - {{{y}}} = {{{0}}}. 


One solution is  {{{y}}} = {{{0}}},  i.e.   {{{cos(2x)}}} = {{{0}}},  which gives   {{{2x}}} = {{{pi/2}}}  or   {{{2x}}} = {{{3pi/2}}}. 

Please complete this case yourself.


Next,  if  {{{y}}} is not zero,  then you can reduce the original equation to the form 


{{{2y}}} - {{{1}}} = {{{0}}},   i.e.   {{{2cos(2x)}}} = {{{1}}}.


It gives you   {{{cos(2x)}}} = {{{1/2}}},  and then   {{{2x}}} = {{{pi/3}}}  or   {{{2x}}} = {{{2pi - pi/3}}} = {{{5pi/3}}}. 


You can complete this case yourself. 



The remaining equations can be solved using the same trick with introduction of a new variable. 


You can do it yourself.


Good luck !!!