Pregunta
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Find the solution for the equation below if \( \beta \in\left(-180^{\circ} ; 180^{\circ}\right] \) : \[ \begin{array}{l}-\sin \left(2 \beta-90^{\circ}\right)+4=3,69\end{array} \]

Ask by Brewer Elliott. in South Africa
Mar 13,2025

Solución de inteligencia artificial de Upstudy

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The solutions for \(\beta\) within \(-180^\circ\) to \(180^\circ\) are approximately: \[ \beta = -125.97^\circ, \quad \beta = -54.03^\circ, \quad \beta = 54.03^\circ, \quad \beta = 125.97^\circ. \]

Solución

We start with the equation \[ -\sin\left(2\beta-90^{\circ}\right)+4=3.69. \] ### Step 1. Isolate the Sine Term Subtract 4 from both sides: \[ -\sin\left(2\beta-90^{\circ}\right)=3.69-4=-0.31. \] Multiply both sides by \(-1\): \[ \sin\left(2\beta-90^{\circ}\right)=0.31. \] ### Step 2. Solve the Sine Equation Let \[ x=2\beta-90^{\circ}. \] Then the equation becomes \[ \sin x=0.31. \] The general solution for \(\sin x=0.31\) is \[ x=\arcsin(0.31)+360^{\circ}k\quad\text{or}\quad x=180^{\circ}-\arcsin(0.31)+360^{\circ}k,\quad k\in \mathbb{Z}. \] We denote \[ \theta_0=\arcsin(0.31) \approx 18.06^{\circ}. \] The two families of solutions for \(x\) are: 1. \(x=18.06^{\circ}+360^{\circ}k\), 2. \(x=180^{\circ}-18.06^{\circ}+360^{\circ}k=161.94^{\circ}+360^{\circ}k\). ### Step 3. Return to \(\beta\) Recall that \[ x=2\beta-90^{\circ}. \] #### First family: \[ 2\beta-90^{\circ} = 18.06^{\circ} + 360^{\circ}k. \] Add \(90^{\circ}\) to both sides: \[ 2\beta =108.06^{\circ} +360^{\circ}k. \] Divide by 2: \[ \beta =54.03^{\circ} +180^{\circ}k. \] #### Second family: \[ 2\beta-90^{\circ} = 161.94^{\circ} +360^{\circ}k. \] Add \(90^{\circ}\): \[ 2\beta =251.94^{\circ} +360^{\circ}k. \] Divide by 2: \[ \beta =125.97^{\circ} +180^{\circ}k. \] ### Step 4. Determine the Solutions in the Given Interval We need the solutions such that \[ \beta\in\left(-180^{\circ},180^{\circ}\right]. \] #### For the first family: \(\beta =54.03^{\circ}+180^{\circ}k\) - For \(k=0\): \(\beta=54.03^{\circ}\) (valid). - For \(k=-1\): \(\beta=54.03^{\circ}-180^{\circ}=-125.97^{\circ}\) (valid). - For \(k=1\): \(\beta=234.03^{\circ}\) (not in the interval). #### For the second family: \(\beta =125.97^{\circ}+180^{\circ}k\) - For \(k=0\): \(\beta=125.97^{\circ}\) (valid). - For \(k=-1\): \(\beta=125.97^{\circ}-180^{\circ}=-54.03^{\circ}\) (valid). - For \(k=1\): \(\beta=305.97^{\circ}\) (not in the interval). ### Final Answer The solutions within \(\beta\in\left(-180^{\circ},180^{\circ}\right]\) are approximately: \[ \beta=-125.97^{\circ},\quad \beta=-54.03^{\circ},\quad \beta=54.03^{\circ},\quad \beta=125.97^{\circ}. \]

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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: \[ \beta \approx 36.17^\circ, \quad -36.17^\circ, \quad -143.83^\circ \]

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