Question 1197640
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Solve for θ. 0° ≤ θ ≤ 360°.
(sin^2)θ+(1/(sin^2)θ)+sinθ+(1/sinθ)=4
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<pre>
Introduce new variable x = {{{sin(theta)}}} + {{{1/sin(theta)}}}.


Notice that {{{x^2}}} = {{{sin^2(theta)}}} + {{{1/sin^2(theta)}}} + 2;

so          {{{sin^2(theta)}}} + {{{1/sin^2(theta)}}} = {{{x^2-2}}}.


THEREFORE, the original equation takes the form

    {{{x^2-2}}} + x = 4,

or

    {{{x^2 +x - 6}}} = 0.


Factor left side

    {x+3)*(x-2) = 0.


The roots are x= -3  and  x= 2.


Next we consider two cases.


    (a)  if x= -3, it means  {{{sin(theta)}}} + {{{1/sin(theta)}}} = -3,  which implies

         {{{sin^2(theta) + 3*sin(theta) + 1}}} = 0,  and then, due to the quadratic formula

         {{{sin(theta)}}} = {{{(-3 +- sqrt(3^2-4*1))/2}}} = {{{(-3 +- sqrt(5))/2}}}.

         It gives only one root  {{{sin(theta)}}} = {{{(-3 + sqrt(5))/2}}} = -0.382  (rounded),
 
         so  {{{theta}}} = -arcsin(0.382) = 360° - 22.457° = 337.543° 
         
         or  {{{theta}}} = 180° + arcsin(0.382) = 180° + 22.457° = 202.457°.



    (b)  if x= 2, it means  {{{sin(theta)}}} + {{{1/sin(theta)}}} = 2,  which implies

         {{{sin^2(theta) - 2*sin(theta) + 1}}} = 0,  and then

         {{{(sin(theta)-1)^2}}} = 0.

         It gives one root  {{{sin(theta)}}} = 1 of multiplicity 2,  so  {{{theta}}} = 90° of multiplicity 2.


<U>ANSWER</U>.  The solutions are  {{{theta}}} = 90° of multiplicity 2,  {{{theta}}} = 202.457°  and  {{{theta}}} = 337.543°.
</pre>

Solved.


The key to the solution is the substitution made at the very beginning of my post.


It is a standard way to solve such equations, but far not everyone knows it.


It is what you need to learn from my solution: how it works.