Problem 45 - Algebra 2

Question

Solve each equation for all values of \(\theta\) \(\tan ^{2} \theta+\sqrt{3}=(1+\sqrt{3}) \tan \theta\)

Step-by-Step Solution

Answer

Solve the quadratic and return roots \( x \); then, \( \theta = \tan^{-1}(x) + n\pi \), for any integer \( n \).

Step 1: Move All Terms to One Side

First, rearrange the equation by moving all terms to one side to set the equation to zero: \[ \tan^2 \theta + \sqrt{3} - (1 + \sqrt{3}) \tan \theta = 0 \]

Step 2: Replace Variables for Simplicity

Let's set \( x = \tan \theta \). Substitute this into the equation: \[ x^2 + \sqrt{3} - (1 + \sqrt{3})x = 0 \]

Step 3: Arrange as a Standard Quadratic Equation

Rearrange the equation to form a standard quadratic equation: \[ x^2 - (1 + \sqrt{3})x + \sqrt{3} = 0 \]

Step 4: Use Quadratic Formula

Apply the quadratic formula, \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1 \), \( b = -(1 + \sqrt{3}) \), and \( c = \sqrt{3} \): \[ x = \frac{1 + \sqrt{3} \pm \sqrt{(1+\sqrt{3})^2 - 4 \cdot 1 \cdot \sqrt{3}}}{2} \]

Step 5: Simplify the Discriminant

Calculate the discriminant part under the square root: \[(1+\sqrt{3})^2 - 4\sqrt{3} = 1 + 2\sqrt{3} + 3 - 4\sqrt{3} = 4 - 2\sqrt{3} \]

Step 6: Solve for x using the Quadratic Formula

Continue simplifying the expression: \[ x = \frac{1 + \sqrt{3} \pm \sqrt{4 - 2\sqrt{3}}}{2} \]Calculate each root separately.

Step 7: Convert Roots Back to \( \theta \)

Using the found values of \( x \), solve for \( \theta \): \[ \theta = \tan^{-1}(x) + n\pi \] where \( n \) is any integer indicating the periodic nature of the tangent function.

Key Concepts

Quadratic Equations

Quadratic equations appear frequently in mathematics, and mastering them is crucial. They are polynomial equations of the second degree, typically expressed in the form \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants. In our problem, the equation is \(x^2 - (1 + \sqrt{3})x + \sqrt{3} = 0\), resembling a quadratic equation when we substitute \(x\) for \(\tan \theta\).

To solve the quadratic equations, we usually use the quadratic formula:

• \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)

This formula is derived from completing the square technique and provides the roots of any quadratic equation.

Quadratic equations can have two solutions, one solution, or no real solutions at all, depending on the discriminant \(b^2 - 4ac\):

• If the discriminant is positive, there are two real solutions.
• If it is zero, there is one real solution.
• If it is negative, there are no real solutions.

Learning to recognize these cases aids in predicting the nature of solutions without fully solving the equation.

Trigonometric Identities

Trigonometric identities are equations involving trigonometric functions that are true for every value of the occurring variables for which both sides of the equation are defined. These identities are essential tools in simplifying expressions and solving trigonometric equations.

Common trigonometric identities include:

• Reciprocal Identities: such as \(\sin \theta = \frac{1}{\csc \theta}\)
• Pythagorean Identities: such as \(\sin^2 \theta + \cos^2 \theta = 1\)
• Angle Sum and Difference Identities: such as \(\sin(a \pm b) = \sin a \cos b \pm \cos a \sin b\)

In this exercise, while the main focus isn't directly on using identities, understanding them can make tackling future trigonometric problems more intuitive. They provide alternative paths and confirm the correctness of solutions by cross-verifying through different methodologies.

Tangent Function

The tangent function, \(\tan \theta\), is one of the fundamental trigonometric functions. It is defined in the context of a right-angled triangle as the ratio of the opposite side to the adjacent side.

More formally, in terms of sine and cosine functions, it is given by:

• \(\tan \theta = \frac{\sin \theta}{\cos \theta}\)

This function has a period of \(\pi\), meaning it repeats every \(\pi\) radians. It's important to remember that tangent can be undefined if \(\cos \theta = 0\) because division by zero is undefined.

Tangent is an odd function, implying:

• \(\tan(-\theta) = -\tan(\theta)\)

This property helps solve equations involving tangent, as it affects how solutions are patterned across different intervals on the angle plane. Knowing how tangent behaves aids in understanding where additional solutions may arise or where to avoid pitfalls in calculations.

Inverse Trigonometric Functions

Inverse trigonometric functions take a certain range of values and return the corresponding angles. The inverse tangent function, written as \(\tan^{-1} x\) or \(\arctan x\), provides the angle whose tangent is \(x\).

This function is critical when transforming back from a calculated trigonometric function value into an angular measure. When a problem states \(\theta = \tan^{-1}(x) + n\pi\), it demonstrates the periodic nature of tangent and the principal value used to represent inverse tangent:

• \(\theta\) typically lies between \(-\frac{\pi}{2}\) and \(\frac{\pi}{2}\) (principal value range).
• Adding \(n\pi\) accounts for the periodic repetition of tangent.

Having a solid grasp of inverse functions allows for reversing calculations, whether it's converting current angles to their trigonometric mean or finding specific angle measures from given trigonometric values.