sin^2x + 0.5tanx = 1

asked by guest
on Nov 16, 2024 at 2:25 am



You asked:

Solve the equation \({\sin\left( x \right)}^{2} + 0.5 \cdot \tan\left( x \right) = 1\) for the variable \(x\).

MathBot Answer:

The real solutions are: \[\begin{aligned}x &= \left\{\frac{\pi \left(5 + 8 n\right)}{4}\; \middle|\; n \in \mathbb{Z}\right\} \approx \left\{6.2831853 n + 3.9269908\; \middle|\; n \in \mathbb{Z}\right\}\\x &= \left\{\frac{\pi \left(1 + 8 n\right)}{4}\; \middle|\; n \in \mathbb{Z}\right\} \approx \left\{6.2831853 n + 0.78539816\; \middle|\; n \in \mathbb{Z}\right\}\end{aligned}\]

The complex solutions are: \[\begin{aligned}x &= \left\{\pi - \arctan{\left(\frac{\sqrt{2 + \sqrt{2}}}{\sqrt{2 - \sqrt{2}}} \right)} + i \left(- \frac{\log{\left(3 + \sqrt{7} \right)}}{2} + \frac{\log{\left(2 \right)}}{4}\right) + 2 \pi n\; \middle|\; n \in \mathbb{Z}\right\} \approx \left\{6.2831853 n + 1.9634954 - 0.69216485 i\; \middle|\; n \in \mathbb{Z}\right\}\\x &= \left\{\pi - \arctan{\left(\frac{\sqrt{2 + \sqrt{2}}}{\sqrt{2 - \sqrt{2}}} \right)} + i \left(- \frac{\log{\left(3 - \sqrt{7} \right)}}{2} + \frac{\log{\left(2 \right)}}{4}\right) + 2 \pi n\; \middle|\; n \in \mathbb{Z}\right\} \approx \left\{6.2831853 n + 1.9634954 + 0.69216485 i\; \middle|\; n \in \mathbb{Z}\right\}\\x &= \left\{- \arctan{\left(\frac{\sqrt{2 + \sqrt{2}}}{\sqrt{2 - \sqrt{2}}} \right)} + 2 \pi + i \left(- \frac{\log{\left(3 + \sqrt{7} \right)}}{2} + \frac{\log{\left(2 \right)}}{4}\right) + 2 \pi n\; \middle|\; n \in \mathbb{Z}\right\} \approx \left\{6.2831853 n + 5.1050881 - 0.69216485 i\; \middle|\; n \in \mathbb{Z}\right\}\\x &= \left\{- \arctan{\left(\frac{\sqrt{2 + \sqrt{2}}}{\sqrt{2 - \sqrt{2}}} \right)} + 2 \pi + i \left(- \frac{\log{\left(3 - \sqrt{7} \right)}}{2} + \frac{\log{\left(2 \right)}}{4}\right) + 2 \pi n\; \middle|\; n \in \mathbb{Z}\right\} \approx \left\{6.2831853 n + 5.1050881 + 0.69216485 i\; \middle|\; n \in \mathbb{Z}\right\}\end{aligned}\]

\(i\) is the imaginary unit, defined as \(i^2 = -1\).

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