Find the solution for the equation below if \( \beta \in\left(-180^{\circ} ; 180^{\circ}\right] \) : \[ \begin{array}{l}-\sin \left(2 \beta-90^{\circ}\right)+4=3,69\end{array} \]
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First, let's simplify the equation: \[ -\sin(2\beta - 90^\circ) + 4 = 3.69 \] Subtract 4 from both sides: \[ -\sin(2\beta - 90^\circ) = 3.69 - 4 \] \[ -\sin(2\beta - 90^\circ) = -0.31 \] Multiply both sides by -1: \[ \sin(2\beta - 90^\circ) = 0.31 \] The sine function can be rewritten using the identity \(\sin(x) = \cos(90^\circ - x)\): \[ \cos(2\beta) = 0.31 \] Now, we need to find the angles that yield this value: \[ 2\beta = \cos^{-1}(0.31) \quad \text{or} \quad 2\beta = -\cos^{-1}(0.31) + 360^\circ k \quad (k \in \mathbb{Z}) \] Calculating \(\cos^{-1}(0.31)\): \[ \cos^{-1}(0.31) \approx 72.34^\circ \] Thus, we have two cases: 1) \[ 2\beta = 72.34^\circ + 360^\circ k \] 2) \[ 2\beta = -72.34^\circ + 360^\circ k \] Now, dividing both equations by 2 gives us: 1) \[ \beta = 36.17^\circ + 180^\circ k \] 2) \[ \beta = -36.17^\circ + 180^\circ k \] Considering \(\beta \in (-180^\circ; 180^\circ]\), let's find valid solutions by substituting possible integer values for k. For \(k = 0\): 1) \[ \beta = 36.17^\circ \] 2) \[ \beta = -36.17^\circ \] For \(k = -1\): 1) \[ \beta = 36.17^\circ - 180^\circ = -143.83^\circ \] 2) \[ \beta = -36.17^\circ - 180^\circ = -216.17^\circ \quad \text{(not valid)} \] For \(k = 1\): 1) \[ \beta = 36.17^\circ + 180^\circ = 216.17^\circ \quad \text{(not valid)} \] Thus, the valid solutions are: \[ \beta \approx 36.17^\circ, \quad \beta \approx -36.17^\circ, \quad \text{and} \quad \beta \approx -143.83^\circ \] Therefore, the final solutions for \(\beta\) within the specified range are: \[ \beta \approx 36.17^\circ, \quad -36.17^\circ, \quad -143.83^\circ \]
