Question 983734


One solution is 


{{{cos(2x)}}} = {{{0}}}. 


It gives   {{{2x}}} = {{{pi/2 + k*pi}}},  where  {{{k}}}  is an arbitrary integer  (k = 0, +/-1, +/-2, . . . ).


Hence  {{{x}}} = {{{pi/4 + k*(pi/2)}}}.


It gives for  k = 0, 1, 2, 3


{{{x}}} = {{{pi/4}}},  {{{pi/4 + pi/2}}} = {{{3pi/4}}},   {{{pi/4 + pi}}} = {{{5pi/4}}},   {{{pi/4 + 3pi/2}}} = {{{7pi/4}}}. 


Next,  if  {{{cos(2x)}}} is not zero,  then you can reduce the original equation to the form 


{{{cos(2x)}}} = {{{1/2}}}. 


dividing it by {{{2cos(2x)}}}.


It gives you 


{{{2x}}} = +/-{{{pi/3}}} + {{{2k*pi}}},  where  {{{k}}}  is an arbitrary integer  (k = 0, +/-1, +/-2, . . . ).


Hence  {{{x}}} = +/-{{{pi/6}}} + {{{k*pi)}}}.


It gives for  k = 0, 1, 2, 3, 


{{{x}}} = {{{pi/6}}},  {{{pi/6 + pi}}} = {{{7pi/6}}},   {{{-pi/6 }}},   {{{-pi/6 + pi}}} = {{{5pi/6}}}. 


<B>Answer</B>. {{{x}}} = {{{pi/4}}}, &nbsp;{{{3pi/4}}},  &nbsp;{{{5pi/4}}},  &nbsp;{{{7pi/4}}}. 


&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;{{{x}}} = {{{pi/6}}}, &nbsp;{{{7pi/6}}}, &nbsp;{{{-pi/6}}}, &nbsp;{{{5pi/6}}}.