(1-sen^2(y))*csc(y)=1
You asked:
MathBot Answer:
The real solutions are: \[\begin{aligned}y &= \left\{2 \pi n + \arctan{\left(\frac{\sqrt{2} \sqrt{-1 + \sqrt{5}}}{2} \right)}\; \middle|\; n \in \mathbb{Z}\right\} \approx \left\{6.2831853 n + 0.66623943\; \middle|\; n \in \mathbb{Z}\right\}\\y &= \left\{\pi - \arctan{\left(\frac{\sqrt{2} \sqrt{-1 + \sqrt{5}}}{2} \right)} + 2 \pi n\; \middle|\; n \in \mathbb{Z}\right\} \approx \left\{6.2831853 n + 2.4753532\; \middle|\; n \in \mathbb{Z}\right\}\end{aligned}\]
The complex solutions are: \[\begin{aligned}y &= \left\{\frac{3 \pi}{2} + i \left(\frac{\log{\left(2 \right)}}{2} - \frac{\log{\left(4 + 2 \sqrt{5} + \sqrt{2} \sqrt{1 + \sqrt{5}} + \sqrt{10} \sqrt{1 + \sqrt{5}} \right)}}{2}\right) + 2 \pi n\; \middle|\; n \in \mathbb{Z}\right\} \approx \left\{6.2831853 n + 4.712389 - 1.0612751 i\; \middle|\; n \in \mathbb{Z}\right\}\\y &= \left\{\frac{3 \pi}{2} + i \left(\frac{\log{\left(2 \right)}}{2} - \frac{\log{\left(4 + 2 \sqrt{5} - \sqrt{2} \sqrt{1 + \sqrt{5}} - \sqrt{10} \sqrt{1 + \sqrt{5}} \right)}}{2}\right) + 2 \pi n\; \middle|\; n \in \mathbb{Z}\right\} \approx \left\{6.2831853 n + 4.712389 + 1.0612751 i\; \middle|\; n \in \mathbb{Z}\right\}\end{aligned}\]
\(i\) is the imaginary unit, defined as \(i^2 = -1\).