🔄 Solving Trig Equations

Master the art of solving trigonometric equations with pattern recognition and systematic approaches.

🎯 Recognition Cues

Look for these patterns:

• Single trig function: $\sin x = a$, $\cos x = b$, $\tan x = c$
• Multiple trig functions: $\sin x + \cos x = 1$
• Squared functions: $\sin^2 x + \cos^2 x = 1$
• Angle relationships: $\sin(2x) = \cos(x)$
• Inverse functions: $\arcsin x = \frac{\pi}{4}$

📋 Template Solution (5 Steps)

1. Isolate the trigonometric function
2. Identify the reference angle
3. Determine quadrants based on sign
4. Find all solutions in given interval
5. Check for extraneous solutions

🔍 Common Patterns

Pattern 1: Single Function

Template: $f(x) = a$ where $f$ is $\sin$, $\cos$, or $\tan$

Example: Solve $\sin x = \frac{1}{2}$ for $0 \leq x < 2\pi$

Solution:

1. Reference angle: $\arcsin\frac{1}{2} = \frac{\pi}{6}$
2. Quadrants: Sine is positive in I and II
3. Solutions: $x = \frac{\pi}{6}$ and $x = \pi - \frac{\pi}{6} = \frac{5\pi}{6}$

Pattern 2: Squared Functions

Template: Use Pythagorean identity $\sin^2 x + \cos^2 x = 1$

Example: Solve $\sin^2 x = \cos^2 x$ for $0 \leq x < 2\pi$

Solution:

1. Use identity: $\sin^2 x = 1 - \sin^2 x$
2. Solve: $2\sin^2 x = 1$ → $\sin^2 x = \frac{1}{2}$ → $\sin x = \pm\frac{\sqrt{2}}{2}$
3. Reference angle: $\frac{\pi}{4}$
4. Solutions: $x = \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4}$

Pattern 3: Multiple Functions

Template: Use identities to convert to single function

Example: Solve $\sin x + \cos x = 1$ for $0 \leq x < 2\pi$

Solution:

1. Square both sides: $(\sin x + \cos x)^2 = 1$
2. Expand: $\sin^2 x + 2\sin x \cos x + \cos^2 x = 1$
3. Use identities: $1 + \sin(2x) = 1$ → $\sin(2x) = 0$
4. Solve: $2x = 0, \pi, 2\pi, 3\pi, \ldots$ → $x = 0, \frac{\pi}{2}, \pi, \frac{3\pi}{2}$
5. Check: Only $x = 0$ and $x = \frac{\pi}{2}$ satisfy original equation

🎯 AMC-Style Worked Example

Problem: Find all real solutions to $\sin(2x) = \cos(x)$ in the interval $[0, 2\pi)$.

Solution:

1. Use double angle formula: $\sin(2x) = 2\sin x \cos x$
2. Substitute: $2\sin x \cos x = \cos x$
3. Factor: $\cos x(2\sin x - 1) = 0$
4. Solve each factor:
   • $\cos x = 0$ → $x = \frac{\pi}{2}, \frac{3\pi}{2}$
   • $2\sin x - 1 = 0$ → $\sin x = \frac{1}{2}$ → $x = \frac{\pi}{6}, \frac{5\pi}{6}$
5. All solutions: $x = \frac{\pi}{6}, \frac{\pi}{2}, \frac{5\pi}{6}, \frac{3\pi}{2}$

⚠️ Common Pitfalls

Pitfall: Forgetting to check all quadrants

Fix: Always consider both positive and negative cases for $\sin$ and $\cos$.

Pitfall: Extraneous solutions from squaring

Fix: Check all solutions in the original equation.

Pitfall: Wrong period for multiple angles

Fix: For $\sin(kx) = a$, period is $\frac{2\pi}{k}$.

Pitfall: Domain restrictions on inverse functions

Fix: Remember ranges: $\arcsin \in [-\frac{\pi}{2}, \frac{\pi}{2}]$, $\arccos \in [0, \pi]$.

📋 Quick Reference

Special Angle Values

Common Identities

• $\sin^2 x + \cos^2 x = 1$
• $\sin(2x) = 2\sin x \cos x$
• $\cos(2x) = \cos^2 x - \sin^2 x = 2\cos^2 x - 1 = 1 - 2\sin^2 x$

Solution Intervals

• For $[0, 2\pi)$: Check all quadrants
• For $[0, \pi)$: Check quadrants I and II only
• For $(-\pi, \pi]$: Check all four quadrants

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