Step-by-step solutions for quadratic trigonometric equations

📚 Quadratic Trigonometric Equations Explained

Quadratic trigonometric equations are equations that combine quadratic expressions with trigonometric functions. They often appear in the form of $a[trig(x)]^2 + b[trig(x)] + c = 0$, where $trig(x)$ represents a trigonometric function like sine, cosine, or tangent, and $a$, $b$, and $c$ are constants. Solving these equations involves using trigonometric identities, algebraic manipulation, and finding solutions within specified intervals.

📜 A Brief History

The development of trigonometric equations is intertwined with the history of trigonometry itself, dating back to ancient civilizations such as the Babylonians and Greeks, who used trigonometric relationships for astronomy and surveying. The formalization of trigonometric functions and their application in equations developed gradually over centuries, with significant contributions from mathematicians like Ptolemy, Aryabhata, and later, European mathematicians during the Renaissance and Enlightenment periods. The concept of quadratic equations can be traced back to Babylonian mathematics.

🔑 Key Principles for Solving

• 🍎 Substitution: Introduce a variable, say $u$, to represent the trigonometric function (e.g., $u = sin(x)$). This transforms the equation into a standard quadratic equation.
• ➗ Solving the Quadratic: Solve the resulting quadratic equation for $u$ using factoring, completing the square, or the quadratic formula.
• 🔄 Back-Substitution: Substitute the trigonometric function back in for $u$ (e.g., $sin(x) = value$).
• 📐 Finding Solutions: Find all angles $x$ that satisfy the trigonometric equation within the given interval (usually $0 \le x < 2\pi$). Use the unit circle or trigonometric tables to find these values.
• ➕ General Solutions: If no interval is specified, find the general solution by adding integer multiples of the period of the trigonometric function (e.g., $2n\pi$ for sine and cosine).

✍️ Step-by-Step Solution

Here's a detailed breakdown of the steps involved:

1. 🔍 Simplify: Begin by simplifying the equation as much as possible using trigonometric identities.
2. 📝 Rewrite: Write the equation in the standard quadratic form: $a[trig(x)]^2 + b[trig(x)] + c = 0$.
3. 💡 Substitute: Let $u = trig(x)$, so the equation becomes $au^2 + bu + c = 0$.
4. ➗ Solve: Solve the quadratic equation for $u$. You can use factoring, completing the square, or the quadratic formula: $u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$
5. 🔄 Back-Substitute: Replace $u$ with $trig(x)$ to get $trig(x) = value_1$ and $trig(x) = value_2$.
6. 📐 Find Solutions: Solve each trigonometric equation for $x$ within the given interval. Use the unit circle, trigonometric tables, or inverse trigonometric functions.
7. ➕ General Solutions (if needed): Write the general solutions by adding integer multiples of the period of the trigonometric function to each solution. For example, if $sin(x) = value$, the general solution is $x = arcsin(value) + 2n\pi$ or $x = \pi - arcsin(value) + 2n\pi$, where $n$ is an integer.

🧮 Example 1: Factoring

Solve $2\sin^2(x) - \sin(x) - 1 = 0$ for $0 \le x < 2\pi$.

1. 🍎 Rewrite: The equation is already in quadratic form.
2. 💡 Substitute: Let $u = \sin(x)$. The equation becomes $2u^2 - u - 1 = 0$.
3. ➗ Solve: Factor the quadratic equation: $(2u + 1)(u - 1) = 0$. So, $u = -\frac{1}{2}$ or $u = 1$.
4. 🔄 Back-Substitute: Replace $u$ with $\sin(x)$: $\sin(x) = -\frac{1}{2}$ or $\sin(x) = 1$.
5. 📐 Find Solutions: For $\sin(x) = -\frac{1}{2}$, $x = \frac{7\pi}{6}$ or $x = \frac{11\pi}{6}$. For $\sin(x) = 1$, $x = \frac{\pi}{2}$.
6. ✅ Final Answer: The solutions are $x = \frac{\pi}{2}, \frac{7\pi}{6}, \frac{11\pi}{6}$.

📈 Example 2: Quadratic Formula

Solve $\cos^2(x) - 3\cos(x) + 1 = 0$ for $0 \le x < 2\pi$.

1. 🍎 Rewrite: The equation is already in quadratic form.
2. 💡 Substitute: Let $u = \cos(x)$. The equation becomes $u^2 - 3u + 1 = 0$.
3. ➗ Solve: Use the quadratic formula: $u = \frac{-(-3) \pm \sqrt{(-3)^2 - 4(1)(1)}}{2(1)} = \frac{3 \pm \sqrt{5}}{2}$. So, $u = \frac{3 + \sqrt{5}}{2}$ or $u = \frac{3 - \sqrt{5}}{2}$.
4. 🔄 Back-Substitute: Replace $u$ with $\cos(x)$: $\cos(x) = \frac{3 + \sqrt{5}}{2}$ or $\cos(x) = \frac{3 - \sqrt{5}}{2}$.
5. 📐 Find Solutions: Since $-1 \le \cos(x) \le 1$, $\cos(x) = \frac{3 + \sqrt{5}}{2}$ is not possible. For $\cos(x) = \frac{3 - \sqrt{5}}{2} \approx 0.382$, $x = arccos(0.382) \approx 1.176$ and $x = 2\pi - arccos(0.382) \approx 5.107$.
6. ✅ Final Answer: The solutions are approximately $x = 1.176, 5.107$.

📝 Practice Quiz

1. ❓Solve $2\cos^2(x) + 3\cos(x) + 1 = 0$ for $0 \le x < 2\pi$.
2. ❓Solve $\tan^2(x) - \tan(x) = 0$ for $0 \le x < \pi$.
3. ❓Solve $4\sin^2(x) - 3 = 0$ for $0 \le x < 2\pi$.
4. ❓Solve $\cos^2(x) - \sin^2(x) = 0$ for $0 \le x < 2\pi$.
5. ❓Solve $2\sin^2(x) - 5\sin(x) + 2 = 0$ for $0 \le x < 2\pi$.
6. ❓Solve $3\tan^2(x) - 1 = 0$ for $0 \le x < \pi$.
7. ❓Solve $2\cos^2(x) + \cos(x) - 1 = 0$ for $0 \le x < 2\pi$.

🌍 Real-World Applications

• 🛰️ Physics: Projectile motion calculations, where angles and velocities are involved.
• 🌊 Engineering: Analyzing oscillating systems and wave phenomena.
• 📈 Navigation: Calculating bearings and distances in surveying and mapping.

🔑 Conclusion

Mastering quadratic trigonometric equations involves understanding the underlying trigonometric identities, applying algebraic techniques for solving quadratic equations, and carefully finding solutions within specified intervals. By following these steps and practicing regularly, you can confidently tackle these types of problems. Good luck! 👍