Solve \( 3 \tan ^{2} \theta-7 \tan \theta+2=0 \) in the interval \( 0^{\circ}<\theta<360^{\circ} \)

To solve the equation \( 3 \tan^2 \theta - 7 \tan \theta + 2 = 0 \), we can use the quadratic formula. Let's denote \( x = \tan \theta \), then the equation becomes: \[ 3x^2 - 7x + 2 = 0. \] Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 3 \), \( b = -7 \), and \( c = 2 \): 1. Calculate the discriminant: \[ b^2 - 4ac = (-7)^2 - 4(3)(2) = 49 - 24 = 25. \] 2. Apply the quadratic formula: \[ x = \frac{7 \pm \sqrt{25}}{2(3)} = \frac{7 \pm 5}{6}. \] 3. This gives us two solutions for \( x \): \[ x_1 = \frac{12}{6} = 2, \quad x_2 = \frac{2}{6} = \frac{1}{3}. \] Now, we need to find \( \theta \) such that \( \tan \theta = 2 \) and \( \tan \theta = \frac{1}{3} \). ### For \( \tan \theta = 2 \): - The reference angle is \( \theta = \tan^{-1}(2) \). This gives one angle in the first quadrant and another in the third quadrant: - \( \theta_1 = \tan^{-1}(2) \approx 63.43^\circ \) - \( \theta_2 = 180^\circ + 63.43^\circ \approx 243.43^\circ \) ### For \( \tan \theta = \frac{1}{3} \): - The reference angle is \( \theta = \tan^{-1}(\frac{1}{3}) \). This also gives one angle in the first quadrant and another in the third quadrant: - \( \theta_3 = \tan^{-1}(\frac{1}{3}) \approx 18.43^\circ \) - \( \theta_4 = 180^\circ + 18.43^\circ \approx 198.43^\circ \) ### Final Solutions: Thus, the solutions for \( \theta \) in the interval \( 0^\circ < \theta < 360^\circ \) are: - \( \theta \approx 63.43^\circ \) - \( \theta \approx 243.43^\circ \) - \( \theta \approx 18.43^\circ \) - \( \theta \approx 198.43^\circ \) So, the four values of \( \theta \) are approximately: - \( 63.43^\circ, 243.43^\circ, 18.43^\circ, 198.43^\circ \).