Solving Trigonometric Equations with the Quadratic Formula: Practical Application in Trigonometry

When dealing with complex trigonometric equations, the quadratic formula can provide a powerful and straightforward method for finding solutions. In this article, we will explore how to apply the quadratic formula to solve equations involving trigonometric functions, specifically focusing on the equation 2cos^2theta - 3sintheta 0.

In general, the standard form of a quadratic equation is given by:

Standard Form of a Quadratic Equation

$$ ax^2 bx c 0 ldots 1 $$

The roots of the equation are given by the quadratic formula:

The Quadratic Formula

$$ x frac{-b pm sqrt{b^2 - 4ac}}{2a} $$

Example Equation: 2cos^2{theta} - 3sin{theta} 0

Given the trigonometric equation:

$$ 2cos^2{theta} - 3sin{theta} 0 $$

We can start by expressing the equation in terms of a new variable, let's say x sintheta. We know that cos^2{theta} 1 - sin^2{theta}. Therefore:

$$ 2(1 - sin^2{theta}) - 3sin{theta} 0 $$

Expanding and simplifying, we get:

$$ 2 - 2sin^2{theta} - 3sin{theta} 0 $$

$$ -2sin^2{theta} - 3sin{theta} 2 0 $$

Multiplying the entire equation by -1 to make the coefficient of the quadratic term positive:

$$ 2sin^2{theta} 3sin{theta} - 2 0 $$

Applying the Quadratic Formula

Now, we can compare the equation 2sin^2{theta} 3sin{theta} - 2 0 with the standard form ax^2 bx c 0. Here, a 2, b 3, and c -2.

Substituting these values into the quadratic formula:

$$ sin{theta} frac{-3 pm sqrt{3^2 - 4(2)(-2)}}{2(2)} $$

$$ sin{theta} frac{-3 pm sqrt{9 16}}{4} $$

$$ sin{theta} frac{-3 pm sqrt{25}}{4} $$

$$ sin{theta} frac{-3 pm 5}{4} $$

Solving for theta

We have two possible solutions for sin{theta}:

Case 1: sin{theta} frac{-3 5}{4} frac{2}{4} frac{1}{2}

However, this value is not valid since the range of the sine function is -1 leq sin{theta} leq 1. Therefore, this solution is not applicable.

Case 2: sin{theta} frac{-3 - 5}{4} frac{-8}{4} -2

This value is also not valid since the range of the sine function is -1 leq sin{theta} leq 1. Therefore, this solution is not applicable either.

Thus, the only valid solution from the quadratic formula is:

$$ sin{theta} -frac{1}{2} $$

The general solutions for theta can be written as:

$$ theta -frac{pi}{6} 2kpi, k in mathbb{Z} $$

and

$$ theta frac{7pi}{6} 2kpi, k in mathbb{Z} $$

Conclusion

In this article, we demonstrated how to use the quadratic formula to solve a complex trigonometric equation. Although the initial quadratic formula resulted in values outside the valid range of the sine function, the application of trigonometric identities and the quadratic formula provided a structured and systematic approach to solving the problem. This method is particularly useful for more complex trigonometric equations and can serve as a valuable tool for students and professionals in mathematics and related fields.