Steps to identify and eliminate extraneous solutions when squaring trig equations

📚 Extraneous Solutions in Trigonometric Equations: A Comprehensive Guide

When solving trigonometric equations, squaring both sides is a common technique. However, this process can introduce extraneous solutions – values that satisfy the transformed equation but not the original one. This guide will walk you through identifying and eliminating these unwanted solutions to ensure accurate results. Let's dive in!

📜 History and Background

The concept of extraneous solutions isn't unique to trigonometry; it arises whenever we perform non-reversible operations on equations. Squaring both sides is one such operation because it doesn't preserve the sign of the expressions involved. This has been understood since the early development of algebra, and mathematicians have devised methods to deal with these pitfalls. Ignoring extraneous solutions can lead to incorrect answers and a misunderstanding of the underlying principles.

🔑 Key Principles

• 📐 Understanding the Source: Squaring both sides eliminates the sign difference. For instance, both $x = 2$ and $x = -2$ satisfy $x^2 = 4$. This is where the extra solutions come from.
• 🧐 The Importance of Verification: Always, always, always substitute your solutions back into the original equation to check if they are valid. This is the golden rule!
• 📈 Considering the Domain: Be mindful of the domain restrictions for trigonometric functions. For example, $\cos^{-1}(x)$ is only defined for $-1 \leq x \leq 1$.

✅ Steps to Identify and Eliminate Extraneous Solutions

• ✍️ Solve the Equation: Start by solving the trigonometric equation as usual, including squaring both sides if necessary.
• 🔄 Isolate the Trig Function: Before squaring, try to isolate the trigonometric function (e.g., $\sin(x)$, $\cos(x)$) on one side of the equation. This can sometimes help minimize the chance of introducing extraneous solutions.
• 🧪 Square Both Sides (If Needed): If squaring is necessary to eliminate radicals or simplify the equation, proceed carefully.
• ✍️ Solve the Resulting Equation: Solve the new trigonometric equation for the variable (e.g., $x$).
• 🔍 Check for Extraneous Solutions: This is the most crucial step. Substitute each potential solution back into the original equation. If the solution does not satisfy the original equation, it is an extraneous solution and must be discarded.
• 💡 Consider the Unit Circle: Use the unit circle to visualize the solutions and identify any values that don't align with the original equation's conditions.
• 📝 Write the Final Solution Set: Only include the solutions that satisfy the original equation.

🌟 Real-World Examples

Example 1:

Solve: $\sqrt{3}\sin(x) = \cos(x)$

1. Square both sides: $3\sin^2(x) = \cos^2(x)$
2. Rewrite using $\cos^2(x) = 1 - \sin^2(x)$: $3\sin^2(x) = 1 - \sin^2(x)$
3. Simplify: $4\sin^2(x) = 1$
4. Solve for $\sin(x)$: $\sin(x) = \pm \frac{1}{2}$
5. Possible solutions: $x = \frac{\pi}{6}, \frac{5\pi}{6}, \frac{7\pi}{6}, \frac{11\pi}{6}$
6. Check in the original equation: Only $x = \frac{\pi}{6}$ and $x = \frac{7\pi}{6}$ are valid.
7. Final solution: $x = \frac{\pi}{6}, \frac{7\pi}{6}$

Example 2:

Solve: $\sin(x) + 1 = \cos(x)$ for $0 \le x < 2\pi$

1. Square both sides: $(\sin(x) + 1)^2 = \cos^2(x)$
2. Expand: $\sin^2(x) + 2\sin(x) + 1 = \cos^2(x)$
3. Use $\cos^2(x) = 1 - \sin^2(x)$: $\sin^2(x) + 2\sin(x) + 1 = 1 - \sin^2(x)$
4. Simplify: $2\sin^2(x) + 2\sin(x) = 0$
5. Factor: $2\sin(x)(\sin(x) + 1) = 0$
6. Solve: $\sin(x) = 0$ or $\sin(x) = -1$
7. Possible solutions: $x = 0, \pi, \frac{3\pi}{2}$
8. Check in the original equation:
• For $x = 0$: $\sin(0) + 1 = 0 + 1 = 1$ and $\cos(0) = 1$. It works!
• For $x = \pi$: $\sin(\pi) + 1 = 0 + 1 = 1$ and $\cos(\pi) = -1$. It does not work!
• For $x = \frac{3\pi}{2}$: $\sin(\frac{3\pi}{2}) + 1 = -1 + 1 = 0$ and $\cos(\frac{3\pi}{2}) = 0$. It works!
9. Final Solution: $x = 0, \frac{3\pi}{2}$

🎯 Conclusion

Squaring trigonometric equations can be a powerful tool, but always remember the importance of checking for extraneous solutions. By carefully verifying each solution against the original equation, you can ensure accurate and reliable results. Happy solving! 🚀