Question 1030398
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{{{3*cos(beta) + 3}}} = {{{2sin^2(beta)}}}.

Replace  {{{sin^2(beta)}}}  by  {{{1 - cos^2(beta)}}}.  You will get

{{{3*cos(beta) + 3}}} = {{{2*(1 - cos^2(beta))}}},   or

{{{3*cos(beta) + 3}}} = {{{2 - 2*cos^2(beta)}}},   or

{{{2*cos^2(beta) + 3*cos(beta) + 1}}} = {{{0}}},   or  (after factoring left side)

{{{(2*cos(beta) + 1)*(cos(beta) + 1)}}} = {{{0}}}.


It gives you two equations.


1)  {{{2*cos(beta) + 1}}} = {{{0}}}  --->  {{{cos(beta)}}} = {{{-1/2}}}  --->  beta = {{{2pi/3 + 2k*pi}}}  or  beta = {{{4pi/3 + 2k*pi}}},  k = 0, +/-1, +/-2, . . . 


2)  {{{cos(beta) + 1}}} = {{{0}}}  --->  {{{cos(beta)}}} = {{{-1}}}  --->  beta = {{{pi + 2k*pi}}} = {{{(2k+1)*pi}}},  k = 0, +/-1, +/-2, . . . 
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