Question 1197640
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{{{sin^2(theta) + 1/(sin^2(theta)) + sin(theta) + 1/(sin(theta)) = 4}}}


{{{m^2 + 1/(m^2) + m + 1/m = 4}}} Let m = sin(theta)


{{{m^4 + 1 + m^3 + m = 4m^2}}} Multiply each term by the LCD m^2 to clear out the fractions.


{{{m^4+m^3+m+1-4m^2 = 0}}}


{{{m^4+m^3-4m^2+m+1 = 0}}}


It appears you followed these steps, or similar to them, since you arrived at what I got shown above. 
The steps are there to help any future student down the road with a similar problem. 


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The task is now solving {{{m^4+m^3-4m^2+m+1 = 0}}} for m.


Use the rational root theorem to find the only possible rational roots are: -1 and 1.


Test each possible root. If you plugged in m = -1, then you should find {{{m^4+m^3-4m^2+m+1}}} turns into {{{(-1)^4+(-1)^3-4(-1)^2+(-1)+1 = -4}}}. This shows m = -1 is NOT a root of the fourth degree polynomial. A root is some input that leads to an output of zero.


Try m = 1 and you should get 0 as a result.
This shows m = 1 is a root of the fourth degree polynomial.
This then means (m-1) is a factor of m^4+m^3-4m^2+m+1


Use either polynomial long division or synthetic division to find that
{{{m^4+m^3-4m^2+m+1 = (m-1)(m^3+2m^2-2m-1)}}}
I skipped showing the steps, but let me know if you need to see them. 


Then use the rational root theorem on {{{m^3+2m^2-2m-1}}} to find that m = 1 is a root of that as well.


It turns out m = 1 is a double root of {{{m^4+m^3-4m^2+m+1}}}


Use polynomial long division or synthetic division to find that
{{{m^3+2m^2-2m-1 = (m-1)(m^2+3m+1)}}}


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Overall, we can say 
{{{m^4+m^3-4m^2+m+1 = (m-1)(m^3+2m^2-2m-1)}}}
{{{m^4+m^3-4m^2+m+1 = (m-1)(m-1)(m^2+3m+1)}}}


Each m-1 factor leads to m = 1 as a root
Which in turn means {{{sin(theta) = 1}}}


Use a chart, calculator, or the unit circle to find the only solution  to {{{sin(theta) = 1}}} is {{{theta = matrix(1,2,90,degrees)}}} where {{{0 <= theta <= 360}}}


You'll need the quadratic formula to compute the roots of {{{m^2+3m+1=0}}}
I'll skip the steps, but you should have these two roots:
{{{m = (-3+sqrt(5))/2}}} and {{{m = (-3-sqrt(5))/2}}}
They approximate to 
{{{(-3+sqrt(5))/2 = -0.381966}}} and {{{(-3-sqrt(5))/2 = -2.618034}}}


Therefore, we have these additional equations to solve for theta
{{{sin(theta) = -0.381966}}} and {{{sin(theta) = -2.618034}}}


But wait, the second equation isn't possible since the range of sine is {{{-1<= sin(theta)<=1}}}. 
The lowest sine can get is -1, which means we can't reach -2.618034 unless we involve complex numbers. 
I'll assume your teacher is working with the real number set only.
In short, we can rule out {{{sin(theta) = -2.618034}}}


I'll leave the task of solving {{{sin(theta) = -0.381966}}} to the student. 
Hint: there are 2 solutions in the interval {{{0 <= theta <= 360}}}; one in quadrant III and the other in quadrant IV
I recommend using the unit circle.


I also recommend graphing software like Desmos or GeoGebra to confirm that you found the correct roots of {{{m^4+m^3-4m^2+m+1}}} (there are 3 distinct roots). Furthermore, that graphing software should also verify each solution for the trig functions in terms of theta.
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