solve tan^2 2x - 13sec2x =1

asked by guest
on Nov 15, 2024 at 10:33 am



You asked:

Solve the equation \({\tan\left( 2 x \right)}^{2} - 13 \cdot \sec\left( 2 x \right) = 1\) for the variable \(x\).

MathBot Answer:

The real solutions are: \[\begin{aligned}x &= \left\{\frac{\arctan{\left(\frac{\sqrt{-165 + 13 \sqrt{177}} \cdot \left(\sqrt{354} + 13 \sqrt{2}\right)}{8} \right)}}{2} + \pi n\; \middle|\; n \in \mathbb{Z}\right\} \approx \left\{3.1415927 n + 0.7473446\; \middle|\; n \in \mathbb{Z}\right\}\\x &= \left\{\pi - \frac{\arctan{\left(\frac{\sqrt{-165 + 13 \sqrt{177}} \cdot \left(\sqrt{354} + 13 \sqrt{2}\right)}{8} \right)}}{2} + \pi n\; \middle|\; n \in \mathbb{Z}\right\} \approx \left\{3.1415927 n + 2.394248\; \middle|\; n \in \mathbb{Z}\right\}\end{aligned}\]

The complex solutions are: \[\begin{aligned}x &= \left\{\frac{i \left(- \log{\left(\frac{13}{4} + \frac{\sqrt{177}}{4} - \frac{\sqrt{2} \sqrt{165 + 13 \sqrt{177}}}{4} \right)} - \pi i \left(1 + 2 n\right)\right)}{2}\; \middle|\; n \in \mathbb{Z}\right\} \approx \left\{0.5 i \left(- 3.1415927 i \left(2.0 n + 1.0\right) + 2.5707471\right)\; \middle|\; n \in \mathbb{Z}\right\}\\x &= \left\{\frac{i \left(- \log{\left(\frac{13}{4} + \frac{\sqrt{177}}{4} + \frac{\sqrt{2} \sqrt{165 + 13 \sqrt{177}}}{4} \right)} - \pi i \left(1 + 2 n\right)\right)}{2}\; \middle|\; n \in \mathbb{Z}\right\} \approx \left\{0.5 i \left(- 3.1415927 i \left(2.0 n + 1.0\right) - 2.5707471\right)\; \middle|\; n \in \mathbb{Z}\right\}\end{aligned}\]

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

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