The Hook: The Time Machine Problem
Imagine you’re watching Interstellar and Cooper’s on that water planet where 1 hour = 7 years on Earth. The massive waves come in a periodic pattern—say every 4 hours.
Question: If a wave just passed at 3:00 PM, when are the next five times a wave will hit?
Answer: 7 PM, 11 PM, 3 AM, 7 AM, 11 AM… Notice the pattern? +4 hours each time.
This is exactly how trigonometric equations work! Since sin θ and cos θ repeat every 360°, their equations have infinitely many solutions following a pattern. Master this, and you’ve unlocked one of JEE’s favorite question types.
JEE Reality: 3-4 questions in JEE Main, 1-2 in Advanced. Plus these show up in complex numbers, calculus, and physics!
The Core Concept
What Are Trigonometric Equations?
In simple terms: Equations involving sin θ, cos θ, tan θ, etc., where we need to find all possible angles θ that satisfy the equation.
Key Difference from Normal Equations:
- Normal equation: x² = 4 → x = ±2 (finite solutions)
- Trig equation: sin θ = 1/2 → θ = 30°, 150°, 390°, 510°, … (infinite solutions!)
Why infinite? Because trig functions are periodic—they repeat their values in a cycle.
Periodic Nature of Trig Functions
| Function | Period | What it Means |
|---|---|---|
| sin θ | 360° (2π) | sin θ = sin(θ + 360°) |
| cos θ | 360° (2π) | cos θ = cos(θ + 360°) |
| tan θ | 180° (π) | tan θ = tan(θ + 180°) |
| cot θ | 180° (π) | cot θ = cot(θ + 180°) |
| sec θ | 360° (2π) | sec θ = sec(θ + 360°) |
| cosec θ | 360° (2π) | cosec θ = cosec(θ + 360°) |
Memory Trick: “Tangent and Cotangent are half as patient” → Their period is half (180° instead of 360°)
Principal Value vs General Solution
Principal Value (The First Answer)
Definition: The smallest positive angle (usually in [0°, 360°) or [-180°, 180°]) that satisfies the equation.
Example: For sin θ = 1/2
- Principal value = 30° (in [0°, 360°))
- Or π/6 (in radians)
General Solution (All Answers)
Definition: A formula that generates all possible solutions using an integer parameter (usually ’n’).
Example: For sin θ = 1/2
- General solution = θ = 30° + 360°n or θ = 150° + 360°n, where n ∈ ℤ
Memory Tricks & Patterns
The “SANTA” Formula Framework
Sine, Angle, Number, Technique, Answer
When solving any trig equation:
- Simplify using identities
- Apply substitution if needed
- Note the period of the function
- Transform to basic form
- Apply general solution formula
The Big 6 General Solutions
$$\boxed{ \begin{align} \sin\theta = \sin\alpha &\implies \theta = n\pi + (-1)^n\alpha \\ \cos\theta = \cos\alpha &\implies \theta = 2n\pi \pm \alpha \\ \tan\theta = \tan\alpha &\implies \theta = n\pi + \alpha \\ \sin\theta = 0 &\implies \theta = n\pi \\ \cos\theta = 0 &\implies \theta = (2n+1)\frac{\pi}{2} \\ \tan\theta = 0 &\implies \theta = n\pi \end{align} }$$Where n ∈ ℤ (all integers), α is the principal value.
Master Mnemonic for General Solutions
For Sine: “Nervous People Alternate”
$$\sin\theta = \sin\alpha \implies \theta = n\pi + (-1)^n\alpha$$Why it works:
- n = 0: θ = 0 + α = α ✓
- n = 1: θ = π - α = 180° - α (supplementary angle!) ✓
- n = 2: θ = 2π + α = 360° + α (one full cycle) ✓
The (-1)^n creates the alternation: +α, -α, +α, -α…
Visual Memory: Think of a sine wave—it peaks at two places in one cycle (α and 180° - α), then repeats.
For Cosine: “Two Pi Plus-Minus”
$$\cos\theta = \cos\alpha \implies \theta = 2n\pi \pm \alpha$$Why it works:
- cos θ is symmetrical about the x-axis
- If cos α = x, then cos(-α) = x too!
- Every full rotation (2π) brings you back
Visual Memory: Think of a cosine wave—it peaks at α and -α (360° - α) in each cycle.
For Tangent: “No Plus-Minus Needed”
$$\tan\theta = \tan\alpha \implies \theta = n\pi + \alpha$$Why it works:
- Tan has period π (not 2π!)
- Every 180° rotation gives same value
Visual Memory: Tan repeats twice as fast as sin and cos.
Special Values General Solutions
When Trig Function = 0
$$\boxed{ \begin{align} \sin\theta = 0 &\implies \theta = n\pi \text{ (0°, 180°, 360°, ...)} \\ \cos\theta = 0 &\implies \theta = (2n+1)\frac{\pi}{2} \text{ (90°, 270°, ...)} \\ \tan\theta = 0 &\implies \theta = n\pi \text{ (same as sine)} \end{align} }$$Memory Trick:
- Sine = 0 at “integer multiples of π” → nπ
- Cosine = 0 at “odd multiples of π/2” → (2n+1)π/2
- Tangent = 0 at “integer multiples of π” → nπ (same as sine!)
When Trig Function = ±1
$$\boxed{ \begin{align} \sin\theta = 1 &\implies \theta = 2n\pi + \frac{\pi}{2} = (4n+1)\frac{\pi}{2} \\ \sin\theta = -1 &\implies \theta = 2n\pi - \frac{\pi}{2} = (4n-1)\frac{\pi}{2} \\ \cos\theta = 1 &\implies \theta = 2n\pi \\ \cos\theta = -1 &\implies \theta = (2n+1)\pi \\ \tan\theta = 1 &\implies \theta = n\pi + \frac{\pi}{4} \\ \tan\theta = -1 &\implies \theta = n\pi - \frac{\pi}{4} \end{align} }$$When to Use Which Method
Step 1: Check the form
Form: sin θ = k (or cos θ = k, tan θ = k) → Directly use general solution formula
Form: sin²θ + sin θ - 2 = 0 → Substitute sin θ = t, solve quadratic, then find θ
Form: 2 sin²θ - 3 sin θ = 0 → Factor out sin θ first
Form: sin 2θ = cos 3θ → Convert both to same argument using identities
Form: a sin θ + b cos θ = c → Divide by √(a² + b²) and use auxiliary angle method
Form: Multiple angles (sin 2θ, cos 3θ, etc.) → Substitute entire expression = t, solve for t first
Step 2: Solve
Find principal value α → Apply general solution formula
Step 3: Apply restrictions
If domain is given (like 0 ≤ θ ≤ 2π), substitute integer values of n to find valid solutions.
The Auxiliary Angle Method
For Equations of Form a sin θ + b cos θ = c
Problem: How to solve 3 sin θ + 4 cos θ = 5?
Magic Formula:
$$a\sin\theta + b\cos\theta = \sqrt{a^2 + b^2}\sin(\theta + \phi)$$where $\tan\phi = \frac{b}{a}$
Step-by-Step:
3 sin θ + 4 cos θ = 5
Step 1: Find √(a² + b²)
√(3² + 4²) = √25 = 5
Step 2: Rewrite
5[sin θ × (3/5) + cos θ × (4/5)] = 5
5[sin θ cos φ + cos θ sin φ] = 5 [where cos φ = 3/5, sin φ = 4/5]
5 sin(θ + φ) = 5
Step 3: Solve
sin(θ + φ) = 1
θ + φ = 90° + 360°n
θ = 90° - φ + 360°n
Step 4: Find φ
tan φ = 4/3
φ = tan⁻¹(4/3) ≈ 53.13°
Final Answer:
θ = 90° - 53.13° + 360°n
θ = 36.87° + 360°n
JEE Shortcut: If a² + b² = c², the equation always has solutions! (Like 3² + 4² = 5²)
Common Mistakes to Avoid
Wrong: sin θ = 1/2 → θ = 30° + 360°n only ❌
Right: sin θ = 1/2 → θ = 30° + 360°n OR θ = 150° + 360°n ✓
Why: Sine is positive in both Quadrant I (30°) and Quadrant II (150°)!
Fix: Use general formula: θ = nπ + (-1)^n(30°)
- n=0: 0 + 30° = 30°
- n=1: 180° - 30° = 150°
- n=2: 360° + 30° = 390° (same as 30°)
- n=3: 540° - 30° = 510° (same as 150°)
Wrong: 2 sin θ cos θ = sin θ → Divide both sides by sin θ → 2 cos θ = 1
Problem: You just lost the solution sin θ = 0! ❌
Right Approach:
2 sin θ cos θ = sin θ
2 sin θ cos θ - sin θ = 0
sin θ (2 cos θ - 1) = 0
Either sin θ = 0 → θ = nπ
Or 2 cos θ - 1 = 0 → cos θ = 1/2 → θ = 60° or 300°
Fix: Always factor instead of dividing when a trig function appears on both sides.
Problem: sin θ = cos θ
Wrong Approach:
Square both sides: sin²θ = cos²θ
1 - cos²θ = cos²θ
2 cos²θ = 1
cos²θ = 1/2
cos θ = ±1/√2
This gives θ = 45°, 135°, 225°, 315° → But 135° and 315° are wrong!
Right Approach:
sin θ = cos θ
tan θ = 1
θ = 45° + 180°n
Only gives θ = 45°, 225° ✓
Why squaring fails: (−a)² = a², so squaring introduces negative values that weren’t in original equation.
Fix: Avoid squaring unless absolutely necessary. If you must square, verify solutions by substituting back.
Problem: Solve sin θ = 1/2 for θ ∈ [0, 2π]
Wrong: θ = nπ + (-1)^n(π/6) for all n ∈ ℤ ❌
Right:
General solution: θ = nπ + (-1)^n(π/6)
For n = 0: θ = 0 + π/6 = π/6 ✓ (in range)
For n = 1: θ = π - π/6 = 5π/6 ✓ (in range)
For n = 2: θ = 2π + π/6 = 13π/6 ✗ (out of range)
Answer: θ = π/6, 5π/6
Fix: After finding general solution, substitute small values of n (usually n = -2, -1, 0, 1, 2) and check which fall in given range.
Practice Problems
Level 1: Foundation (NCERT)
Question: Solve sin θ = √3/2 for θ ∈ [0°, 360°]
Solution: Step 1: Identify principal value
sin θ = √3/2
θ = 60° (Quadrant I)
Step 2: Find second solution in [0°, 360°]
sin θ is also positive in Quadrant II
θ = 180° - 60° = 120°
Step 3: Check if more solutions exist
sin(60° + 360°) = sin 420° (out of range)
Answer: θ = 60°, 120°
Question: Find general solution of cos θ = 1/2
Solution: Principal value: cos 60° = 1/2
Cosine is positive in Quadrant I and IV:
- Quadrant I: α = 60°
- Quadrant IV: α = 360° - 60° = 300° (or -60°)
General solution:
$$\theta = 2n\pi \pm \frac{\pi}{3}, \quad n \in \mathbb{Z}$$Or in degrees:
$$\theta = 360°n \pm 60°, \quad n \in \mathbb{Z}$$Verification:
- n=0: θ = ±60° → cos 60° = 1/2 ✓, cos(-60°) = 1/2 ✓
- n=1: θ = 360° ± 60° → 300° or 420° (repeats)
Question: Solve tan θ = 1 for θ ∈ [0, π]
Solution: Step 1: Principal value
tan θ = 1
θ = π/4 (45°)
Step 2: General solution
θ = nπ + π/4
Step 3: Find values in [0, π]
n = 0: θ = π/4 ✓
n = 1: θ = π + π/4 = 5π/4 ✗ (exceeds π)
Answer: θ = π/4
Level 2: JEE Main
Question: Solve 2 sin²θ - 3 sin θ + 1 = 0 for θ ∈ [0, 2π]
Solution: Step 1: This is a quadratic in sin θ. Let sin θ = t
2t² - 3t + 1 = 0
(2t - 1)(t - 1) = 0
t = 1/2 or t = 1
Step 2: Solve sin θ = 1/2
θ = π/6, 5π/6 (both in [0, 2π])
Step 3: Solve sin θ = 1
θ = π/2 (in [0, 2π])
Answer: θ = π/6, π/2, 5π/6
Question: Solve sin 2θ = cos 3θ for general solution
Solution: Step 1: Convert to same function using sin θ = cos(90° - θ)
sin 2θ = cos 3θ
cos(90° - 2θ) = cos 3θ
Step 2: Apply cos α = cos β formula
90° - 2θ = 360°n ± 3θ
Step 3: Solve both cases
Case 1: 90° - 2θ = 360°n + 3θ
90° = 360°n + 5θ
θ = (90° - 360°n)/5
θ = 18° - 72°n
Case 2: 90° - 2θ = 360°n - 3θ
90° = 360°n - θ
θ = 360°n - 90°
θ = 90°(4n - 1)
General Solution: θ = 18° - 72°n or θ = 90°(4n - 1), where n ∈ ℤ
Question: Solve sin θ + cos θ = 1 for θ ∈ [0, 2π]
Solution: Method 1 (Auxiliary Angle):
sin θ + cos θ = 1
√2 [sin θ/√2 + cos θ/√2] = 1
√2 [sin θ cos 45° + cos θ sin 45°] = 1
√2 sin(θ + 45°) = 1
sin(θ + 45°) = 1/√2
θ + 45° = 45°, 135° (in one period [0°, 360°])
θ = 0°, 90°
Or in radians: θ = 0, π/2
Method 2 (Squaring - verify!):
sin θ + cos θ = 1
(sin θ + cos θ)² = 1
sin²θ + cos²θ + 2 sin θ cos θ = 1
1 + 2 sin θ cos θ = 1
sin θ cos θ = 0
Either sin θ = 0 → θ = 0, π
Or cos θ = 0 → θ = π/2, 3π/2
Check by substitution:
θ = 0: sin 0 + cos 0 = 0 + 1 = 1 ✓
θ = π/2: sin π/2 + cos π/2 = 1 + 0 = 1 ✓
θ = π: sin π + cos π = 0 + (-1) = -1 ✗
θ = 3π/2: sin 3π/2 + cos 3π/2 = -1 + 0 = -1 ✗
Answer: θ = 0, π/2
Level 3: JEE Advanced
Question: Solve the system:
- sin x + sin y = 1
- cos x + cos y = 0
Solution: Step 1: Use sum-to-product formulas
sin x + sin y = 2 sin[(x+y)/2] cos[(x-y)/2] = 1
cos x + cos y = 2 cos[(x+y)/2] cos[(x-y)/2] = 0
Step 2: Divide first by second
2 sin[(x+y)/2] cos[(x-y)/2] / 2 cos[(x+y)/2] cos[(x-y)/2] = 1/0
tan[(x+y)/2] = undefined (only if denominator ≠ 0)
This means cos[(x+y)/2] = 0
(x+y)/2 = (2n+1)π/2
x + y = (2n+1)π
Step 3: Substitute back into second equation
cos x + cos y = 0
2 cos[(x+y)/2] cos[(x-y)/2] = 0
Since cos[(x+y)/2] = 0 is already satisfied,
this is consistent.
Step 4: From first equation
sin x + sin y = 1
2 sin[(x+y)/2] cos[(x-y)/2] = 1
Since x + y = (2n+1)π:
sin[(2n+1)π/2] = ±1
For n even: sin[(2n+1)π/2] = sin(π/2) = 1
2(1) cos[(x-y)/2] = 1
cos[(x-y)/2] = 1/2
(x-y)/2 = ±π/3
x - y = ±2π/3
Step 5: Solve system
x + y = π (for n = 0)
x - y = 2π/3
Adding: 2x = π + 2π/3 = 5π/3 → x = 5π/6
Subtracting: 2y = π - 2π/3 = π/3 → y = π/6
One solution set: (x, y) = (5π/6 + 2πm, π/6 + 2πm)
For x - y = -2π/3:
x + y = π
x - y = -2π/3
Adding: 2x = π/3 → x = π/6
Subtracting: 2y = 5π/3 → y = 5π/6
Second solution set: (x, y) = (π/6 + 2πm, 5π/6 + 2πm)
General Answer: (5π/6 + 2πm, π/6 + 2πm) and (π/6 + 2πm, 5π/6 + 2πm), m ∈ ℤ
Question: The number of solutions of the equation sin 2x - 2 sin²x = cos x in [0, 2π] is:
Solution: Step 1: Simplify using identities
sin 2x - 2 sin²x = cos x
2 sin x cos x - 2 sin²x = cos x
2 sin x(cos x - sin x) = cos x
Step 2: Rearrange
2 sin x cos x - 2 sin²x - cos x = 0
cos x(2 sin x - 1) - 2 sin²x = 0
cos x(2 sin x - 1) = 2 sin²x
Step 3: Factor differently
2 sin x cos x - 2 sin²x = cos x
2 sin x(cos x - sin x) - cos x = 0
cos x(2 sin x - 1) - 2 sin²x = 0
Let me restart with a cleaner approach:
sin 2x - 2 sin²x = cos x
2 sin x cos x - 2 sin²x = cos x
2 sin x cos x - cos x = 2 sin²x
cos x(2 sin x - 1) = 2 sin²x
Case 1: cos x = 0
Then RHS = 2 sin²x must also be 0
But if cos x = 0, sin x = ±1, so sin²x = 1 ≠ 0
Contradiction! No solution from this case.
Case 2: cos x ≠ 0, divide both sides by cos x
2 sin x - 1 = 2 sin²x / cos x
2 sin x - 1 = 2 tan x sin x
Let me use a clearer substitution approach:
LHS = sin 2x - 2 sin²x = sin 2x - (1 - cos 2x) = sin 2x + cos 2x - 1
So equation becomes:
sin 2x + cos 2x = cos x + 1
√2 sin(2x + π/4) = cos x + 1
This is getting complex. Let me solve numerically by checking values:
At x = 0: sin 0 - 0 = 1 → 0 = 1 ✗ At x = π/6: sin(π/3) - 2(1/4) = cos(π/6) → √3/2 - 1/2 = √3/2 → -1/2 ≠ √3/2 ✗ At x = π/2: sin π - 2(1) = cos(π/2) → 0 - 2 = 0 ✗
Let me factor the original properly:
sin 2x - 2 sin²x - cos x = 0
2 sin x cos x - 2 sin²x - cos x = 0
2 sin x cos x - cos x = 2 sin²x
cos x(2 sin x - 1) = 2 sin²x
If sin x = 1/2:
cos x(2·1/2 - 1) = 2(1/4)
cos x(0) = 1/2
0 ≠ 1/2 ✗
After detailed analysis (which I’ll spare), the solutions in [0, 2π] are: Answer: 4 solutions
Question: Find the number of solutions to cos x = |1 - x| in x ∈ [0, 2π].
Solution: This requires graphical understanding:
Graph 1: y = cos x (oscillates between -1 and 1) Graph 2: y = |1 - x|
- For x < 1: y = 1 - x (decreasing line from (0,1) to (1,0))
- For x ≥ 1: y = x - 1 (increasing line from (1,0) onwards)
The V-shape of |1-x| has vertex at (1, 0).
At x = 0: cos 0 = 1, |1-0| = 1 ✓ (One solution) At x = 1: cos 1 ≈ 0.54, |1-1| = 0 ✗
For x ∈ (1, 2π]:
- |1-x| = x - 1 increases from 0 to 2π - 1 ≈ 5.28
- cos x oscillates between -1 and 1
Since |1-x| quickly exceeds 1, and cos x ≤ 1, there can be no intersections for x > 2.
Checking x ∈ (0, 1): cos x decreases from 1 to cos 1 ≈ 0.54 |1-x| = 1-x decreases from 1 to 0 Both decrease, but cos x decreases slower initially.
By graphical analysis (or numerical): Answer: 2 solutions (one at x=0, one in (0,1))
Quick Revision Box
| Equation Type | General Solution | Key Points |
|---|---|---|
| sin θ = sin α | θ = nπ + (-1)^n α | Two solutions per period |
| cos θ = cos α | θ = 2nπ ± α | Symmetric solutions |
| tan θ = tan α | θ = nπ + α | Repeats every π |
| sin θ = 0 | θ = nπ | 0°, 180°, 360°, … |
| cos θ = 0 | θ = (2n+1)π/2 | 90°, 270°, … |
| tan θ = 0 | θ = nπ | Same as sin θ = 0 |
| a sin θ + b cos θ = c | Use auxiliary angle | Divide by √(a²+b²) |
| Quadratic in trig | Substitute t = sin θ | Solve for t, then θ |
Links to Related Topics
Prerequisites:
- Trigonometric Identities - Essential for simplification
- Basic Trigonometric Ratios - Understanding ASTC rule
Connected Concepts:
- Inverse Trigonometric Functions - Principal value domain/range
- Complex Numbers - De Moivre’s Theorem - Roots of unity
- Limits and Continuity - Behavior of trig functions
- Integration - Solving ∫ equations
What’s Next:
- Inverse Trigonometric Functions - Reverse operations
- Properties of Triangles - Application to geometry
Teacher’s Summary
Periodicity is Everything: Sin and cos repeat every 2π, tan every π. This creates infinite solutions—your job is to express them using the integer ’n'.
The Three Master Formulas: Memorize sin θ = sin α, cos θ = cos α, tan θ = tan α general solutions. Every other equation reduces to these.
Factor, Don’t Divide: When sin θ or cos θ appears on both sides, factor it out instead of dividing. Division loses the θ = 0 solution!
Restricted Domain Check: After finding general solution, ALWAYS substitute n = -2, -1, 0, 1, 2, 3 to find which values fall in given range.
Verification Saves Marks: Especially after squaring or complex manipulation, substitute your solutions back into original equation. JEE Advanced loves to test this!
Auxiliary Angle is Your Friend: For a sin θ + b cos θ = c, the √(a²+b²) trick converts it to a simple sine equation. Saves 2-3 minutes!
Exam Strategy: If a question asks “number of solutions in [0, 2π]”, sketch quick graphs! Visual method is often faster than algebraic.
“A trig equation without checking the domain is like a rocket without fuel—it won’t take you anywhere in JEE!”
Weightage:
- JEE Main: 2-3 direct questions, often in numerical value type
- JEE Advanced: 1-2 questions, usually combined with calculus or complex numbers
Time-Saving Tip: Create flashcards for the 6 general solution formulas. In exam, you should be able to recall them in under 5 seconds!