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Find the general solution for this equation: \[ -\tan \left(2 x-45^{\circ}\right)+2=5 \] INSTRUCTION: Round your answers to two decimal places, and if there is more than o the answers with the \( \because{ }^{\circ} \) symbol. Answer: \( x=\square ; k \in \mathbb{Z} \)

Ask by Carlson Carrillo. in South Africa
Mar 12,2025

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Tutor-Verified Answer

Answer

The general solution is \( x = -13.29^\circ + 90^\circ\, k \), where \( k \) is any integer.

Solution

We start with the equation \[ -\tan\left(2x - 45^\circ\right) + 2 = 5. \] **Step 1.** Isolate the tangent term by subtracting 2 from both sides: \[ -\tan\left(2x - 45^\circ\right) = 3. \] Multiply both sides by \(-1\): \[ \tan\left(2x - 45^\circ\right) = -3. \] **Step 2.** Write the general solution for the tangent equation. Recall that if \[ \tan\theta = -3, \] then \[ \theta = \arctan(-3) + 180^\circ \, k, \quad k \in \mathbb{Z}. \] In our case, \(\theta = 2x - 45^\circ\), so: \[ 2x - 45^\circ = \arctan(-3) + 180^\circ\, k, \quad k \in \mathbb{Z}. \] **Step 3.** Evaluate \(\arctan(-3)\) in degrees (rounding to two decimal places). We have: \[ \arctan(-3) \approx -71.57^\circ. \] Thus, \[ 2x - 45^\circ = -71.57^\circ + 180^\circ\, k. \] **Step 4.** Solve for \(x\) by adding \(45^\circ\) to both sides: \[ 2x = -71.57^\circ + 45^\circ + 180^\circ\, k, \] which simplifies to: \[ 2x = -26.57^\circ + 180^\circ\, k. \] Now divide the equation by 2: \[ x = -13.29^\circ + 90^\circ\, k. \] Thus, the general solution rounded to two decimal places is: \[ \boxed{ x = -13.29^\circ + 90^\circ\, k, \quad k \in \mathbb{Z} }. \]

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To solve the equation \(-\tan(2x - 45^\circ) + 2 = 5\), we first isolate the tangent function: \[ -\tan(2x - 45^\circ) = 3 \] \[ \tan(2x - 45^\circ) = -3 \] Now, we can find the general solution for \(\tan(\theta) = -3\): \[ 2x - 45^\circ = \tan^{-1}(-3) + k \cdot 180^\circ \] Calculating \(\tan^{-1}(-3)\): Using a calculator, \(\tan^{-1}(-3) \approx -71.57^\circ\). Thus, we have: \[ 2x - 45^\circ = -71.57^\circ + k \cdot 180^\circ \] Now, solve for \(x\): \[ 2x = -71.57^\circ + 45^\circ + k \cdot 180^\circ \] \[ 2x = -26.57^\circ + k \cdot 180^\circ \] \[ x = -13.285^\circ + \frac{k \cdot 180^\circ}{2} \] \[ x = -13.29^\circ + 90^\circ k \] In rounding to two decimal places, we get: \[ x \approx -13.29^\circ + 90^\circ k \] Therefore, the general solution is: \[ x = -13.29^\circ + 90^\circ k; \quad k \in \mathbb{Z} \]

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