2sin(x + 15°)= 1,462

asked by guest
on Nov 21, 2024 at 9:47 am



You asked:

Solve the equation \(2 \cdot \sin\left( x + 15^{\circ} \right) = 1,462\) for the variable \(x\).

MathBot Answer:

There are no real solutions.

The complex solutions are: \[\begin{aligned}x &= \left\{2 \pi n - \frac{i \log{\left(1068721 - 2924 \sqrt{133590} \right)}}{2} + \arctan{\left(\frac{- 731 \sqrt{2} - 731 \sqrt{6} + 4 \sqrt{66795} + 12 \sqrt{22265}}{- 731 \sqrt{6} - 4 \sqrt{66795} + 12 \sqrt{22265} + 731 \sqrt{2}} \right)}\; \middle|\; n \in \mathbb{Z}\right\} \approx \left\{6.2831853 n + 1.3089969 + 7.2875602 i\; \middle|\; n \in \mathbb{Z}\right\}\\x &= \left\{- \arctan{\left(\frac{4 \sqrt{66795} + 12 \sqrt{22265} + 731 \sqrt{2} + 731 \sqrt{6}}{- 731 \sqrt{6} - 12 \sqrt{22265} + 4 \sqrt{66795} + 731 \sqrt{2}} \right)} + 2 \pi n - \frac{i \log{\left(1068721 + 2924 \sqrt{133590} \right)}}{2}\; \middle|\; n \in \mathbb{Z}\right\} \approx \left\{6.2831853 n + 1.3089969 - 7.2875602 i\; \middle|\; n \in \mathbb{Z}\right\}\end{aligned}\]

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

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