How to Solve Trigonometric Equations

How to Solve Trigonometric Equations

(For SSC, Bank, Railway Exams – Ace Your Exam with Confidence!)

Introduction

"Mastering trigonometric equations can get you 5-7 marks in SSC CGL, Bank PO, or Railway exams—enough to push you from Tier-2 to Tier-1 or from rejection to selection. These questions look scary, but follow this 5-step method, and you’ll solve them faster than the clock ticks!

What You Need To Know First

Before diving in, ensure you’re solid on: 1. Basic trigonometric ratios (sin, cos, tan) – Know their values at standard angles (0°, 30°, 45°, 60°, 90°). 2. Trigonometric identities – Especially Pythagorean identities (sin²θ + cos²θ = 1) and angle sum/difference formulas. 3. General solutions for sin, cos, tan – How to write all possible angles that satisfy an equation (e.g., sinθ = ½ → θ = 30° + 360°n or 150° + 360°n).

Key Vocabulary

Formulas To Know

Step-by-Step Method

Follow these 5 steps for EVERY trigonometric equation:

1. Isolate the trig function – Get sinθ, cosθ, or tanθ alone on one side.

2. Example: 2sinθ + 1 = 0 → sinθ = -½.

3. Find the principal solution – Use inverse trig functions to find the smallest positive angle (0° to 360°).

4. For sinθ = -½, principal solutions are 210° and 330° (since sin is negative in Quadrants III and IV).

5. Write the general solution – Add 360°n (for sin/cos) or 180°n (for tan) to the principal solutions.

6. For sinθ = -½: θ = 210° + 360°n or 330° + 360°n.

7. Check for restrictions – If the question specifies a range (e.g., 0° ≤ θ ≤ 180°), list only the angles that fit.

8. For 0° ≤ θ ≤ 180°, only 210° is invalid (since 330° > 180°). No solution in this range.

9. Simplify using identities (if needed) – If the equation has sin²θ, cos²θ, or multiple angles, use identities to rewrite it.

10. Example: 2sin²θ + sinθ – 1 = 0 → Let x = sinθ → 2x² + x – 1 = 0 → Solve quadratic.

Worked Example (Using the Steps)

Solve: 2cosθ + √3 = 0, 0° ≤ θ ≤ 360°.

Step 1: Isolate cosθ. → 2cosθ = -√3 → cosθ = -√3/2.

Step 2: Find principal solutions. → cos⁻¹(-√3/2) = 150° (since cos is negative in Quadrants II and III). → Second solution: 360° – 150° = 210°.

Step 3: Write general solution (not needed here since range is given). → θ = 150° + 360°n or 210° + 360°n.

Step 4: Check range (0° ≤ θ ≤ 360°). → Valid solutions: 150° and 210°.

Step 5: No identities needed here.

Final Answer: θ = 150° or 210°.

What we did and why: - We isolated cosθ to make it a standard equation. - Found principal solutions using inverse cosine and the unit circle. - Checked the range to discard extra angles. - No identities were needed because the equation was already simple.

Worked Examples

Example 1 – Basic

Solve: sinθ = ½, 0° ≤ θ ≤ 360°.

Step 1: Already isolated (sinθ = ½).

Step 2: Principal solutions: → sin⁻¹(½) = 30° (Quadrant I). → Second solution: 180° – 30° = 150° (Quadrant II).

Step 3: General solution (not needed here).

Step 4: Range is 0° ≤ θ ≤ 360° → Both 30° and 150° are valid.

Final Answer: θ = 30° or 150°.

What we did and why: - Used the unit circle to find where sinθ = ½. - Remembered that sine is positive in Quadrants I and II.

Example 2 – Medium

Solve: 2sin²θ – sinθ – 1 = 0, 0° ≤ θ ≤ 360°.

Step 1: Let x = sinθ → 2x² – x – 1 = 0.

Step 2: Solve quadratic: → x = [1 ± √(1 + 8)] / 4 = [1 ± 3]/4. → x = 1 or x = -½.

Step 3: Replace x with sinθ: → sinθ = 1 → θ = 90° (only solution in 0° to 360°). → sinθ = -½ → θ = 210° or 330°.

Step 4: Check range → All angles are valid.

Final Answer: θ = 90°, 210°, or 330°.

What we did and why: - Used substitution to turn the equation into a quadratic. - Solved for sinθ = 1 and sinθ = -½ separately. - Remembered that sinθ = 1 has only one solution in 0° to 360°.

Example 3 – Exam-Style

Solve: √3 tanθ + 1 = 0, -180° ≤ θ ≤ 180°.

Step 1: Isolate tanθ. → √3 tanθ = -1 → tanθ = -1/√3.

Step 2: Principal solution: → tan⁻¹(-1/√3) = -30° (since tan is negative in Quadrants II and IV).

Step 3: General solution: → θ = -30° + 180°n.

Step 4: Find solutions in -180° ≤ θ ≤ 180°: → For n = 0: θ = -30°. → For n = 1: θ = -30° + 180° = 150°. → For n = -1: θ = -30° – 180° = -210° (invalid, outside range).

Final Answer: θ = -30° or 150°.

What we did and why: - Isolated tanθ and used inverse tangent. - Remembered that tan has a period of 180°, so we added 180°n. - Checked the range carefully to avoid extra solutions.

Common Mistakes

Exam Traps

1-Minute Recap

(Spoken naturally, as if to a student the night before the exam.)

"Listen up—this is your 60-second crash course for trigonometric equations. First, isolate the trig function—get sinθ, cosθ, or tanθ alone. Second, find the principal solutions using inverse trig (remember: sin and cos have two solutions in 0° to 360°; tan has one). Third, write the general solution—add 360°n for sin/cos, 180°n for tan. Fourth, check the range—if the question says 0° to 180°, don’t give 210°! Fifth, use identities if you see sin²θ, cos²θ, or double angles—simplify before solving. Common mistakes? Forgetting the second solution, mixing up quadrants, or ignoring the range. Exam traps? Disguised equations (like 2sinθcosθ = 1) and negative signs. You’ve got this—practice 3-4 problems tonight, and you’ll solve them in under 2 minutes tomorrow. Good luck!