Solve \( 7 \tan 2 \theta=6 \) for \( 0^{\circ} \leq \theta<360 \)
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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: \[ \theta \approx 20.3^\circ, \quad 110.3^\circ, \quad 200.3^\circ \]
