Solving Trigonometric Equation Using Quadratic Formula and Inverse Trigonometric Functions
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.
