Question - Solving Trigonometric Equation Using Quadratic Formula and Inverse Trigonometric Functions
Solution:
The provided image contains a trigonometric equation and a hint on how to solve it:\[\text{Given: } (\tan(x))^2 - 1.3\tan(x) - 0.48 = 0\]The hint suggests using the quadratic formula, inverse trigonometric functions, and the symmetry of the unit circle to solve the equation.Let's denote $$ \tan(x) = y $$, which transforms the equation into a quadratic form:\[ y^2 - 1.3y - 0.48 = 0 \]Now, we can solve for $$ y $$ using the quadratic formula where $$ a = 1, b = -1.3, $$ and $$ c = -0.48 $$:\[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]Applying the values:\[ y = \frac{1.3 \pm \sqrt{(1.3)^2 - 4(1)(-0.48)}}{2} \]Solve under the square root:\[ y = \frac{1.3 \pm \sqrt{1.69 + 1.92}}{2} \]\[ y = \frac{1.3 \pm \sqrt{3.61}}{2} \]\[ y = \frac{1.3 \pm 1.9}{2} \]This gives us two possible solutions for $$ y $$:\[ y_1 = \frac{1.3 + 1.9}{2} = \frac{3.2}{2} = 1.6 \]\[ y_2 = \frac{1.3 - 1.9}{2} = \frac{-0.6}{2} = -0.3 \]Now, let's turn back to the original variable $$ x $$:For $$ y_1 = \tan(x) = 1.6 $$:\[ x = \arctan(1.6) \]This needs to be solved in the given interval $$[0, 2\pi]$$. Since tangent has a period of $$ \pi $$, there will be two solutions for $$ x $$ in the given interval. To find the second solution, we add $$ \pi $$ to the first solution:\[ x_1 = \arctan(1.6) \]\[ x_2 = \arctan(1.6) + \pi \]For $$ y_2 = \tan(x) = -0.3 $$:\[ x = \arctan(-0.3) \]Similarly, there will be two solutions for $$ x $$ in the interval $$[0, 2\pi]$$:\[ x_3 = \arctan(-0.3) \]\[ x_4 = \arctan(-0.3) + \pi \]Now, let's find the numerical values for these four $$ x $$ solutions. Remember, these will be approximate values as $$ \arctan $$ doesn't usually produce neat numbers:\[ x_1 = \arctan(1.6) \]\[ x_2 = \arctan(1.6) + \pi \]\[ x_3 = \arctan(-0.3) \]\[ x_4 = \arctan(-0.3) + \pi \]Using a calculator (or any computational tool):\[ x_1 \approx 1.012197011 \quad(\text{in radians}) \]\[ x_2 \approx 1.012197011 + \pi \approx 4.153789664 \]\[ x_3 \approx -0.291456794 \quad(\text{in radians}) \]\[ x_4 \approx -0.291456794 + \pi \approx 2.850135858 \]Note: $$ x_3 $$ is negative so we need to add $$ 2\pi $$ to it to bring it within the $$[0, 2\pi]$$ interval:\[ x_3 = -0.291456794 + 2\pi \approx 5.992728452 \]The calculator is showing an approximation for the solution $$ x $$ which looks like:\[ x \approx 0.9828, \arctan(-0.3) + \pi \]We can interpret this as follows:The first solution $$ x_1 $$ is approximately $$ 0.9828 $$ (which could be a rounding discrepancy from the more accurate value we computed, $$ \approx 1.012197011 $$).The second solution $$ x_3 $$ is approximately equal to $$ \arctan(-0.3) + \pi $$, which we computed as $$ \approx 2.850135858 $$ radians.So, we find $$ x $$ to have the following approximate solutions in the interval $$ [0, 2\pi] $$:\[ x_1 \approx 1.012197011 \quad (\text{radians}) \]\[ x_2 \approx 4.153789664 \quad (\text{radians}) \]\[ x_3 \approx 5.992728452 \quad (\text{radians}) \]\[ x_4 \approx 2.850135858 \quad (\text{radians}) \]These solutions correspond to the angles where the original equation holds true within the given interval.
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