Question 275801
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 5\cos^2\Theta\ -\ 4\sin\Theta\ -\ 4\ =\ 0]


Remember


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \cos^2\phi\ +\ \sin^2\phi\ =\ 1]


Therefore


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \cos^2\phi\ =\ 1\ -\ \sin^2\phi\]


So substitute


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 5\left(1\ -\ \sin^2\Theta\right)\ -\ 4\sin\Theta\ -\ 4\ =\ 0]


Simplify:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 5\ -\ 5\sin^2\Theta\ -\ 4\sin\Theta\ -\ 4\ =\ 0]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ -\ 5\sin^2\Theta\ -\ 4\sin\Theta\ +\ 1\ =\ 0]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 5\sin^2\Theta\ +\ 4\sin\Theta\ -\ 1\ =\ 0]


Let *[tex \Large x] represent *[tex \Large \sin\Theta]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 5x^2\ +\ 4x\ -\ 1\ =\ 0]


Factor the quadratic:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(5x\ -\ 1\right)\left(x\ +\ 1\right)\ =\ 0]


Hence,


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ 0.2]


or


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ -1]


But since *[tex \Large x\ =\ \sin\Theta]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sin\Theta\ =\ 0.2]


or


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sin\Theta\ =\ -1]


And finally, considering the interval *[tex \Large 0\ \leq\ \Theta\ \leq\ 2\pi]:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \Theta\ =\ \sin^{-1}(0.2)\ \approx\ 0.2013] or *[tex \LARGE \approx\ \pi\ -\ 0.2013]


and


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \Theta\ =\ \sin^{-1}(-1)\ =\ \frac{3\pi}{2}]



John
*[tex \LARGE e^{i\pi} + 1 = 0]
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