Question 1204440
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{{{sin^2(x)}}} + {{{1/sin^2(x)}}} + sin(x) + {{{1/sin(x)}}} = 4.
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<pre>
Introduce new variable

    y = sin(x) + {{{1/sin(x)}}}.


Then y^2 = sin^2(x) + 2 + 1/sin^2(x),  and the original equation takes the form

    y^2 - 2 + y = 4,

or

    y^2 + y - 6 = 0.


Its roots are (using the quadratic formula)

    {{{y[1,2]}}} = {{{(-1 +- sqrt (1^2 - 4*1*(-6)))/2}}} = {{{(-1 +- sqrt(25))/2}}} = {{{(-1 +- 5)/2}}},

    or  {{{y[1]}}} = {{{(-1+5)/2}}} = 2,  {{{y[2]}}} = {{{(-1-5)/2}}} = -3.


It means that  sin(x) + {{{1/sin(x)}}}  is  EITHER  2  OR  -3.


So, from this point, we have two cases for x.



(1)  sin(x) + {{{1/sin(x)}}} = 2,  which implies

     sin^2(x) - 2sin(x) + 1 = 0,

     (sin(x) - 1)^2 = 0

      sin(x) = 1

          x  = 90 degrees (or  x = {{{pi/2}}}).

      Thus this case is complete.



(2)  sin(x) + {{{1/sin(x)}}} = -3,  which implies

     sin^2(x) + 3*sin(x) + 1 = 0,

     sin(x) = use the quadratic formula = {{{((-3) +- sqrt (3^2 - 4*1*1))/2}}} = {{{(-3 +- sqrt(5))/2}}}. 

     {{{sin(x)[1_]}}} = {{{(-3 + sqrt(5))/2}}} = ~ -0.382  --->  from it, you may find two values for x between 0 and 360 degrees.

     {{{sin(x)[2_]}}} = {{{(-3 - sqrt(5))/2}}} = ~ -2.618  --->  there is no solution for real x. 

      Thus this case is complete, too.
</pre>

Solved.


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Comparing with the solution by tutor @math_tutor2020, the substitution which I used, leads 
to equation/equations of degree 2, ONLY.  There is no need to work with equation of degree 4.