Question 1150592

{{{cos^2(x)+3cos(x)-1=0}}}

Let {{{cos(x )=u}}}

{{{u^2+3u-1=0}}}


For a quadratic equation of the form {{{ax^2+bx+c=0}}} the solutions are:

 {{{x[1,2]=(-b+-sqrt(b^2-4ac))/2a}}}

in your case {{{a=1}}}, {{{b=3}}}, {{{c=-1}}} 

 {{{u[1,2]=(-3+-sqrt(3^2-4*1(-1)))/2*1}}}

{{{u[1,2]=(-3+-sqrt(9+4))/2}}}

{{{u[1,2]=(-3+-sqrt(13))/2}}}

{{{u[1]=(-3+sqrt(13))/2 }}} and {{{u[2]=(-3-sqrt(13))/2}}}

Substitute back {{{u= cos  (x )}}}:

{{{cos  (x )=(-3+sqrt(13))/2}}}=> {{{x= arccos  ((-3+sqrt(13))/2)+2pi*n}}} , {{{x= 2pi- arccos  ((-3+sqrt(13))/2)+2pi*n}}}

or
{{{cos  (x )=(-3-sqrt(13))/2}}}=> {{{none}}}, so, disregard it


solutions in decimal form
{{{x=1.26319  +2 pi *n}}}
{{{x=2pi -1.26319 +2pi* n}}}