Question 1011164
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{{{12*sin^2(theta)}}} = {{{9}}}.
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<pre>
Divide both sides by 12. You will get

{{{sin^2(theta)}}} = {{{3/4}}}.

Take square roots from both sides. You will get

{{{sin(theta)}}} = +/- {{{sqrt(3)/2}}}.

{{{sin(theta)}}} = {{{sqrt(3)/2}}}  -----> {{{theta}}} = +/- {{{pi/3}}} + {{{2*k*pi}}}, k = 0, +/-1, +/-2, . . .   or   {{{theta}}} = {{{2pi/3}}} + {{{2*k*pi}}}, k = 0, +/-1, +/-2, . . . 

{{{sin(theta)}}} = {{{-sqrt(3)/2}}}  -----> {{{theta}}} = {{{-pi/3}}} + {{{2*k*pi}}}, k = 0, +/-1, +/-2, . . .   or   {{{theta}}} = {{{-2pi/3}}} + {{{2*k*pi}}}, k = 0, +/-1, +/-2, . . . 

<U>Answer</U>. The solutions are

         {{{theta}}} = {{{pi/3}}} + {{{2*k*pi}}}, k = 0, +/-1, +/-2, . . .

         {{{theta}}} = {{{2pi/3}}} + {{{2*k*pi}}}, k = 0, +/-1, +/-2, . . .

         {{{theta}}} = {{{-pi/3}}} + {{{2*k*pi}}}, k = 0, +/-1, +/-2, . . . 

         {{{theta}}} = {{{-2pi/3}}} + {{{2*k*pi}}}, k = 0, +/-1, +/-2, . . . 
</pre>