Principal Solutions

• Solutions that lie in the interval \( [0, 2\pi) \) or \( [0°, 360°) \)

• These are the primary or basic solutions

• Typically limited to one or two complete cycles of the function

Example: For \( \sin x = \frac{1}{2} \), principal solutions are \( x = \frac{\pi}{6}, \frac{5\pi}{6} \)

General Solutions

• All possible solutions including those from multiple periods

• Contains the integer \( n \) representing any integer value

• Accounts for the periodic nature of trigonometric functions

Example: For \( \sin x = \frac{1}{2} \), general solution is \( x = n\pi + (-1)^n \frac{\pi}{6} \), where \( n \in \mathbb{Z} \)

Method 1: Algebraic Manipulation

• Isolate the trigonometric function

• Use inverse trigonometric functions to find the angle

Example: \( 2\sin x = 1 \) → \( \sin x = \frac{1}{2} \) → \( x = \frac{\pi}{6} \) or \( \frac{5\pi}{6} \) (in \( [0, 2\pi) \))

Method 2: Factoring

• Factor equations when possible

• Set each factor equal to zero

Example: \( 2\sin^2 x + 3\sin x + 1 = 0 \) → \( (2\sin x + 1)(\sin x + 1) = 0 \)

Method 3: Using Trigonometric Identities

• Apply Pythagorean identities: \( \sin^2 x + \cos^2 x = 1 \)

• Use double angle formulas, sum/difference formulas

• Convert all terms to a single trigonometric function

Method 4: Substitution

• Use substitution for quadratic-type equations

Example: For \( \cos^2 x - 3\cos x + 2 = 0 \), let \( u = \cos x \)

Method 5: Using the CAST Diagram

• Determine which quadrants have positive values

• Cosine positive in quadrants I & IV

• All functions positive in quadrant I

• Sine positive in quadrants I & II

• Tangent positive in quadrants I & III

Example 1: Basic Equation

Problem: Solve \( 2\sin x - 1 = 0 \) for \( x \in [0, 2\pi) \)

Solution:

Step 1: Isolate the trigonometric function

\( 2\sin x = 1 \)

\( \sin x = \frac{1}{2} \)

Step 2: Find angles where sine equals 1/2

Reference angle: \( x = \frac{\pi}{6} \) (30°)

Sine is positive in quadrants I and II

Solutions: \( x = \frac{\pi}{6} \) and \( x = \pi - \frac{\pi}{6} = \frac{5\pi}{6} \)

Example 2: Quadratic Form

Problem: Solve \( 2\sin^2 x + 3\sin x + 1 = 0 \) for \( x \in [0, 2\pi) \)

Solution:

Step 1: Factor the quadratic expression

\( (2\sin x + 1)(\sin x + 1) = 0 \)

Step 2: Set each factor to zero

\( 2\sin x + 1 = 0 \) → \( \sin x = -\frac{1}{2} \)

\( \sin x + 1 = 0 \) → \( \sin x = -1 \)

Step 3: Find solutions for each case

For \( \sin x = -\frac{1}{2} \): \( x = \frac{7\pi}{6}, \frac{11\pi}{6} \) (quadrants III & IV)

For \( \sin x = -1 \): \( x = \frac{3\pi}{2} \)

Solutions: \( x = \frac{7\pi}{6}, \frac{3\pi}{2}, \frac{11\pi}{6} \)

Example 3: Using Identity

Problem: Solve \( 3\cos \theta + 3 = 2\sin^2 \theta \) for \( \theta \in [0, 2\pi) \)

Solution:

Step 1: Use the identity \( \sin^2 \theta = 1 - \cos^2 \theta \)

\( 3\cos \theta + 3 = 2(1 - \cos^2 \theta) \)

\( 3\cos \theta + 3 = 2 - 2\cos^2 \theta \)

Step 2: Rearrange into standard form

\( 2\cos^2 \theta + 3\cos \theta + 1 = 0 \)

Step 3: Factor

\( (2\cos \theta + 1)(\cos \theta + 1) = 0 \)

Step 4: Solve each factor

\( \cos \theta = -\frac{1}{2} \) → \( \theta = \frac{2\pi}{3}, \frac{4\pi}{3} \)

\( \cos \theta = -1 \) → \( \theta = \pi \)

Solutions: \( \theta = \frac{2\pi}{3}, \pi, \frac{4\pi}{3} \)

Example 4: General Solution

Problem: Find the general solution of \( \tan x = \sqrt{3} \)

Solution:

Step 1: Find the principal value

Reference angle: \( \alpha = \frac{\pi}{3} \) (since \( \tan \frac{\pi}{3} = \sqrt{3} \))

Step 2: Apply the general solution formula

For \( \tan \theta = \tan \alpha \), general solution is \( \theta = n\pi + \alpha \)

General Solution: \( x = n\pi + \frac{\pi}{3} \), where \( n \in \mathbb{Z} \)

💡 Periodicity is Key

Sine and cosine have period \( 2\pi \), tangent has period \( \pi \). This creates infinite solutions for most trigonometric equations.

💡 Check Your Domain

Always verify solutions fall within the specified interval. Extraneous solutions may arise from algebraic manipulation.

💡 Reference Angles

The reference angle is always positive and less than 90°. Use it with quadrant analysis to find all solutions.

💡 Unit Circle Mastery

Know the unit circle values for common angles (30°, 45°, 60°) to quickly identify solutions.

💡 Multiple Solutions

Quadratic trigonometric equations can have up to four solutions in \( [0, 2\pi) \), unlike regular quadratics which have at most two.

💡 Degrees vs Radians

For degrees: add 360° to find more solutions. For radians: add \( 2\pi \) for sine/cosine, \( \pi \) for tangent.

💡 Identity Toolkit

Keep Pythagorean identities, double angle formulas, and sum/difference formulas readily available for conversions.

💡 Curriculum Coverage

Trigonometric equations appear in IB Math (SL & HL), AP Precalculus, A-Level Mathematics, GCSE/IGCSE Additional Mathematics, and SAT Math Level 2.

❌ Forgetting All Solutions

Don't stop after finding one solution. Trigonometric functions are periodic, so check all quadrants.

❌ Dividing by Trigonometric Functions

Never divide both sides by \( \sin x \), \( \cos x \), etc., as you might lose solutions where the function equals zero. Factor instead!

❌ Incorrect Quadrant Analysis

Remember CAST diagram: Cosine positive in I & IV, All in I, Sine in I & II, Tangent in I & III.

❌ Misapplying Identities

Double-check identity applications. For example, \( \sin^2 x + \cos^2 x = 1 \), not \( \sin x + \cos x = 1 \).

❌ Calculator Mode Errors

Ensure your calculator is in the correct mode (degrees or radians) matching your problem.

Problem 1

Solve \( 2\cos x + \sqrt{3} = 0 \) for \( x \in [0, 2\pi) \).

Problem 2

Solve \( \tan^2 x - 3 = 0 \) for \( x \in [0, 2\pi) \).