Problem 20 - Algebra 2 and Trigonometry

Question

In \(15-20\) , find, to the nearest hundredth of a radian, the values of \(\theta\) in the interval \(0 \leq \theta<2 \pi\) that satisfy the equation. $$ 2 \cot ^{2} \theta-13 \cot \theta+6=0 $$

Step-by-Step Solution

Answer

The solutions are \(\theta \approx 0.17, 3.31, 1.11, 4.25\) radians.

Step 1: Substitute cotangent with a variable

Let's set \( x = \cot \theta \). This allows us to simplify the equation \(2 \cot^2 \theta - 13 \cot \theta + 6 = 0\) into a standard quadratic equation: \(2x^2 - 13x + 6 = 0\).

Step 2: Solve the quadratic equation

The quadratic equation \(2x^2 - 13x + 6 = 0\) can be solved using the quadratic formula: \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 2\), \(b = -13\), and \(c = 6\).

Step 3: Calculate the discriminant

Calculate the discriminant \(b^2 - 4ac\): \(-13^2 - 4 \times 2 \times 6 = 169 - 48 = 121\). The discriminant is 121.

Step 4: Find roots using the quadratic formula

Substitute the discriminant and other values into the quadratic formula:\(x = \frac{13 \pm \sqrt{121}}{4}\). This gives us \(x = \frac{13 \pm 11}{4}\).

Step 5: Compute the two solutions for x

Solve each part of \(x = \frac{13 \pm 11}{4}\): \(x = \frac{24}{4} = 6\) and \(x = \frac{2}{4} = 0.5\). So, \(x = 6\) or \(x = 0.5\).

Step 6: Find \(\theta\) for each x value, x=6

Since \(x = \cot \theta\), we find \(\theta\) when \(\cot \theta = 6\). \(\theta = \cot^{-1}(6)\) yields principal value \(\theta \approx 0.1651\) radians. However, \(\cot \theta\) is positive in the first and third quadrants, so other possible \(\theta = \pi + 0.1651 \approx 3.3067\) radians.

Step 7: Find \(\theta\) for each x value, x=0.5

Now find \(\theta\) for \(\cot \theta = 0.5\). Obtain \(\theta = \cot^{-1}(0.5)\), with principal value \(\theta \approx 1.1071\) radians. Since \(\cot \theta\) is also positive in the first and third quadrants, another solution is \(\theta = \pi + 1.1071 \approx 4.2487\) radians.

Step 8: Express solutions in nearest hundredth

Round each solution to the nearest hundredth of a radian, giving:\(\theta \approx 0.17\), \(3.31\), \(1.11\), and \(4.25\) radians.

Key Concepts

Quadratic Equations

Quadratic equations are fundamental in algebra and are expressed in the form \( ax^2 + bx + c = 0 \). They contain three components:

• \(a\), the coefficient of \(x^2\)
• \(b\), the coefficient of \(x\)
• \(c\), the constant term

Understanding quadratic equations is akin to solving puzzles where we need to find the values of \(x\) that satisfy the equation.
The solutions for \(x\) are often called "roots".
The most common method to find these roots is the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]The term under the square root, \(b^2 - 4ac\), is known as the discriminant.
The discriminant helps us determine the nature of the roots:

• If it is positive, the equation has two distinct real roots.
• If zero, there is exactly one real root.
• If negative, there are no real roots, only complex ones.

In our example, the equation is \(2x^2 - 13x + 6 = 0\). After calculating the discriminant, we find it to be 121, which is positive. Therefore, we can expect two distinct real solutions for \(x\): 6 and 0.5.

Cotangent Function

The cotangent function is one of the reciprocal trigonometric functions.
It is the reciprocal of the tangent function and is defined as:\[ \cot \theta = \frac{1}{\tan \theta} = \frac{\cos \theta}{\sin \theta} \]Compared to sine and cosine, cotangent has its own set of properties and graph shape.
It is periodic, just like all trigonometric functions, with a period of \(\pi\) radians.The cotangent function is undefined whenever \(\theta\) is a multiple of \(\pi\), as the sine of these angles is zero, making the fraction undefined.
Cotangent is positive in the first and third quadrants and negative in the second and fourth.To solve problems involving \(\cot \theta\), we sometimes substitute \(\cot \theta\) with a variable, such as \(x\).
This allows us to convert complex trigonometric equations into more straightforward algebraic ones.
For example, converting the original equation to \(2x^2 - 13x + 6 = 0\) makes it solvable using the quadratic formula. Once a solution for \(x\) is found, we reverse the substitution: \(\theta = \cot^{-1}(x)\).

Inverse Trigonometric Functions

Inverse trigonometric functions help us find angles that correspond to specific values of trigonometric functions.
For instance, if we know \( y = \cot \theta \), we can find \( \theta \) using the inverse cotangent function, denoted as \( \cot^{-1}(y) \).The inverse trigonometric functions produce angles within specific ranges to ensure they are "principal" values.
For \( \cot^{-1}(x) \), the principal value usually lies between \(0\) and \(\pi\) radians.However, because trigonometric functions can have the same value at different angles, it's essential to understand their behavior throughout different quadrants.

• For positive \(x\) values, \(\cot^{-1}(x)\) gives angles in both the first and third quadrants.
• This means, other than finding the principal value, angles like \(\pi + \cot^{-1}(x)\) also solve the cotangent equation.

In our exercise, we used \(\cot^{-1}(6)\) and \(\cot^{-1}(0.5)\) to determine the specific values of \(\theta\).
Simplifying these expressions and accounting for periodicity, we found solutions between 0 and \(2\pi\) that satisfy the given trigonometric equation.