b^2\left[\frac{\pi }{4}-\:arccos\left(\frac{15\sqrt{2}}{2b}\right)\right]+\left(\frac{15\sqrt{2}}{2}\right)\sqrt{b^2-\frac{225}{2}}=101.25

asked by guest
on Mar 16, 2025 at 5:45 am



You asked:

Solve the equation b2(π4−arccos⁡(1522b))+1522b2−2252=101.25{b}^{2} \left(\frac{\pi}{4} - \arccos\left( \frac{15 \sqrt{2}}{2 b} \right)\right) + \frac{15 \sqrt{2}}{2} \sqrt{{b}^{2} - \frac{225}{2}} = 101.25 for the variable bb.

MathBot Answer:

The complex solutions are: {b  |  b∈R∧16b4arccos⁡2(1522b)−8b4πarccos⁡(1522b)+b4π2+3240b2arccos⁡(1522b)−810b2π−1800b2+366525=0}∖{0}\left\{b\; \middle|\; b \in \mathbb{R} \wedge 16 b^{4} \arccos^{2}{\left(\frac{15 \sqrt{2}}{2 b} \right)} - 8 b^{4} \pi \arccos{\left(\frac{15 \sqrt{2}}{2 b} \right)} + b^{4} \pi^{2} + 3240 b^{2} \arccos{\left(\frac{15 \sqrt{2}}{2 b} \right)} - 810 b^{2} \pi - 1800 b^{2} + 366525 = 0 \right\} \setminus \left\{0\right\}

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