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Solve \( 7 \tan 2 \theta=6 \) for \( 0^{\circ} \leq \theta<360 \)

Ask by Wells Ray. in Singapore
Jan 24,2025

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Solución

Solve the equation by following steps: - step0: Solve for \(\theta\): \(7\tan\left(2\theta \right)=6\) - step1: Find the domain: \(7\tan\left(2\theta \right)=6,\theta \neq \frac{\pi }{4}+\frac{k\pi }{2},k \in \mathbb{Z}\) - step2: Multiply both sides: \(7\tan\left(2\theta \right)\times \frac{1}{7}=6\times \frac{1}{7}\) - step3: Calculate: \(\tan\left(2\theta \right)=6\times \frac{1}{7}\) - step4: Calculate: \(\tan\left(2\theta \right)=\frac{6}{7}\) - step5: Use the inverse trigonometric function: \(2\theta =\arctan\left(\frac{6}{7}\right)\) - step6: Add the period: \(2\theta =\arctan\left(\frac{6}{7}\right)+k\pi ,k \in \mathbb{Z}\) - step7: Solve the equation: \(\theta =\frac{\arctan\left(\frac{6}{7}\right)}{2}+\frac{k\pi }{2},k \in \mathbb{Z}\) - step8: Check if the solution is in the defined range: \(\theta =\frac{\arctan\left(\frac{6}{7}\right)}{2}+\frac{k\pi }{2},k \in \mathbb{Z},\theta \neq \frac{\pi }{4}+\frac{k\pi }{2},k \in \mathbb{Z}\) - step9: Find the intersection: \(\theta =\frac{\arctan\left(\frac{6}{7}\right)}{2}+\frac{k\pi }{2},k \in \mathbb{Z}\) Calculate or simplify the expression \( \arctan\left(\frac{6}{7}\right) \). Calculate the value by following steps: - step0: Calculate: \(\arctan\left(\frac{6}{7}\right)\) Calculate or simplify the expression \( \frac{1}{2} \cdot 0.708626 \). Calculate the value by following steps: - step0: Calculate: \(\frac{1}{2}\times 0.708626\) - step1: Convert the expressions: \(\frac{1}{2}\times \frac{354313}{500000}\) - step2: Multiply the numbers: \(\frac{354313}{1000000}\) Calculate or simplify the expression \( \frac{1}{2} \cdot 0.708626 + \pi \). Calculate the value by following steps: - step0: Calculate: \(\frac{1}{2}\times 0.708626+\pi \) - step1: Multiply the numbers: \(\frac{354313}{1000000}+\pi \) - step2: Reduce fractions to a common denominator: \(\frac{354313}{1000000}+\frac{\pi \times 1000000}{1000000}\) - step3: Transform the expression: \(\frac{354313+\pi \times 1000000}{1000000}\) - step4: Reorder the terms: \(\frac{354313+1000000\pi }{1000000}\) Calculate or simplify the expression \( \frac{1}{2} \cdot 0.708626 + \frac{\pi}{2} \). Calculate the value by following steps: - step0: Calculate: \(\frac{1}{2}\times 0.708626+\frac{\pi }{2}\) - step1: Multiply the numbers: \(\frac{354313}{1000000}+\frac{\pi }{2}\) - step2: Reduce fractions to a common denominator: \(\frac{354313}{1000000}+\frac{\pi \times 500000}{2\times 500000}\) - step3: Multiply the numbers: \(\frac{354313}{1000000}+\frac{\pi \times 500000}{1000000}\) - step4: Transform the expression: \(\frac{354313+\pi \times 500000}{1000000}\) - step5: Reorder the terms: \(\frac{354313+500000\pi }{1000000}\)

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Extra Insights

To solve the equation \( 7 \tan 2\theta = 6 \), we can start by isolating \( \tan 2\theta \): \[ \tan 2\theta = \frac{6}{7} \] Next, we need to find the angles for which this tangent value is true. The tangent function is periodic with a period of \( 180^\circ \), so we will find the reference angle first. Using a calculator or inverse tangent function: \[ 2\theta = \tan^{-1}\left(\frac{6}{7}\right) \] Calculating this, we get: \[ 2\theta \approx 40.6^\circ \] Since the tangent function is also positive in the third quadrant, we have another solution: \[ 2\theta = 180^\circ + 40.6^\circ \approx 220.6^\circ \] Now, we can write two equations for \( \theta \): 1. \( 2\theta = 40.6^\circ \) Therefore, \( \theta = \frac{40.6^\circ}{2} \approx 20.3^\circ \) 2. \( 2\theta = 220.6^\circ \) Therefore, \( \theta = \frac{220.6^\circ}{2} \approx 110.3^\circ \) Now, since the tangent function has a period of \( 180^\circ \), we add \( 180^\circ \) to \( 2\theta \) for additional solutions: 3. \( 2\theta = 40.6^\circ + 180^\circ = 220.6^\circ \) Therefore, \( \theta = \frac{220.6^\circ}{2} \approx 110.3^\circ \) (which we've already found). 4. \( 2\theta = 220.6^\circ + 180^\circ = 400.6^\circ \) Therefore, \( \theta = \frac{400.6^\circ}{2} = 200.3^\circ \) To summarize, the solutions for \( \theta \) in the range \( 0^\circ \leq \theta < 360^\circ \) are: \[ \theta \approx 20.3^\circ, \quad 110.3^\circ, \quad 200.3^\circ \] Thus, the final answer is: \[ \theta \approx 20.3^\circ, \quad 110.3^\circ, \quad 200.3^\circ \]

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