Problem 50 - Precalculus Essentials

Question

Involve trigonometric equations quadratic in form. Solve each equation on the interval \([0,2 \pi)\) $$3 \tan ^{2} x-9=0$$

Step-by-Step Solution

Answer

The solutions for the given trigonometric equation in the interval \([0, 2 \pi)\) are \(x = \pi / 3,4 \pi / 3,2 \pi / 3,5 \pi / 3\).

Step 1: Rearrange the Equation

Start by rearranging the given trigonometric equation to match it with the standard form. The equation can be rewritten as \(3 \tan^2x = 9\). Then, solve for \(\tan^2x\) by dividing each side by 3. Hence, \(\tan^2x = 3\).

Step 2: Square Root Each Side

Take the square root of each side to solve for \(\tan x\). When solving for \(x\), remember to account for both the positive and negative roots. Hence, \(\tan x = \sqrt{3}\) or \(\tan x = -\sqrt{3}\).

Step 3: Apply the Tangent Function

Now apply the tangent function to solve for \(x\) based on the unit circle. The tangent function has a period of \(\pi\). If \(\tan x = \sqrt{3}\), then \(x = \pi / 3,4 \pi / 3\); if \(\tan x = -\sqrt{3}\), then \(x = 2 \pi / 3,5 \pi / 3\) by checking quadrant where tangent is negative. All these solutions are within the given interval.

Key Concepts

Tangent Function

The tangent function, often abbreviated as tan, is one of the primary trigonometric functions. It relates an angle of a right triangle to the ratio of the opposite side over the adjacent side. This function is periodic and repeats its values over specific intervals.

It is defined as:

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

Due to its definition, the tangent function can have undefined values wherever the cosine function is zero, leading to vertical asymptotes in its graph. This typically occurs at angles like \( \frac{\pi}{2}, \frac{3\pi}{2} \), etc.
Beyond its definition, the tangent function has a period of \( \pi \). This means that every \( \pi \) units, the function returns to its starting value. Hence, when solving equations involving \( \tan(x) \), we only consider solutions within one full cycle or within a desired interval, as done in the original exercise. This periodicity is crucial in finding all solutions in a given range, as it helps encapsulate all possible angles that satisfy the given equation.

Quadratic Form

In mathematics, many trigonometric equations are given in quadratic form, especially when involving functions like sine, cosine, and tangent. Recognizing a quadratic form in trigonometric equations can simplify solving processes by borrowing techniques from algebra.

A basic quadratic equation takes the form:

• \( ax^2 + bx + c = 0 \)

Similarly, a trigonometric equation in quadratic form might look like \( a(\tan x)^2 + b(\tan x) + c = 0 \), using tangent as an example.
To solve these, we can apply standard algebraic techniques: rearranging the equation, factoring, or using the quadratic formula. In the exercise, the original equation \( 3 \tan^2 x - 9 = 0 \) is already almost in a quadratic format, which allows us to conveniently solve by isolating \( \tan^2 x \). Recognizing this quadratic structure is essential, as it allows us to address more complex trigonometric problems efficiently.

Unit Circle

The Unit Circle is a fundamental tool in trigonometry that helps visualize and solve equations involving trigonometric functions, like sine, cosine, and tangent. It is a circle with a radius of 1, centered at the origin of the coordinate plane.

On the unit circle, every point \((x, y)\) represents:

• \( x = \cos(\theta) \)
• \( y = \sin(\theta) \)

For tangent, the unit circle provides a geometrical interpretation: the tangent of an angle is the y-coordinate divided by the x-coordinate (or \( \tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} \)).
Solutions for tangent involve identifying angles where this division yields the values computed from the trigonometric equation. For example, if \( \tan(x) = \sqrt{3} \), we search for angles on the unit circle where \( \frac{y}{x} = \sqrt{3} \).
The unit circle becomes particularly useful when determining which quadrants a solution lies in, especially since the sign of the tangent changes between quadrants. This understanding is essential when finding multiple solutions for the tangent over an interval, as seen in the original exercise where solutions spanned different quadrants.