3cotx=8secx

asked by guest
on Nov 18, 2024 at 2:24 am



You asked:

Solve the equation \(3 \cdot \cot\left( x \right) = 8 \cdot \sec\left( x \right)\) for the variable \(x\).

MathBot Answer:

The real solutions are: \[\begin{aligned}x &= \left\{2 \pi n + \arctan{\left(\frac{\sqrt{2}}{4} \right)}\; \middle|\; n \in \mathbb{Z}\right\} \approx \left\{6.2831853 n + 0.33983691\; \middle|\; n \in \mathbb{Z}\right\}\\x &= \left\{\pi - \arctan{\left(\frac{\sqrt{2}}{4} \right)} + 2 \pi n\; \middle|\; n \in \mathbb{Z}\right\} \approx \left\{6.2831853 n + 2.8017557\; \middle|\; n \in \mathbb{Z}\right\}\end{aligned}\]

The complex solutions are: \[\begin{aligned}x &= \left\{\frac{3 \pi}{2} - i \log{\left(3 + 2 \sqrt{2} \right)} + 2 \pi n\; \middle|\; n \in \mathbb{Z}\right\} \approx \left\{6.2831853 n + 4.712389 - 1.7627472 i\; \middle|\; n \in \mathbb{Z}\right\}\\x &= \left\{\frac{3 \pi}{2} - i \log{\left(3 - 2 \sqrt{2} \right)} + 2 \pi n\; \middle|\; n \in \mathbb{Z}\right\} \approx \left\{6.2831853 n + 4.712389 + 1.7627472 i\; \middle|\; n \in \mathbb{Z}\right\}\end{aligned}\]

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

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