JEE MATH APEX
JEE MATH APEX
JEE MATH APEX
JEE MATH APEX
JEE MATH APEX
JEE MATH APEX
JEE MATH APEX
JEE MATH APEX
TRIGONOMETRIC EQUATIONS
— JEE MATH APEX —
JEE MATH APEX
1. Basic General Solutions
📌 Fundamental Trigonometric Equations
• sin θ = 0 ⇒ θ = nπ, n ∈ Z
• cos θ = 0 ⇒ θ = (2n+1)π/2, n ∈ Z
• tan θ = 0 ⇒ θ = nπ, n ∈ Z
• cot θ = 0 ⇒ θ = (2n+1)π/2, n ∈ Z
• sec θ = 1 ⇒ θ = 2nπ, n ∈ Z
• cosec θ = 1 ⇒ θ = (4n+1)π/2, n ∈ Z
• cos θ = 0 ⇒ θ = (2n+1)π/2, n ∈ Z
• tan θ = 0 ⇒ θ = nπ, n ∈ Z
• cot θ = 0 ⇒ θ = (2n+1)π/2, n ∈ Z
• sec θ = 1 ⇒ θ = 2nπ, n ∈ Z
• cosec θ = 1 ⇒ θ = (4n+1)π/2, n ∈ Z
JEE MATH APEX
2. General Solutions with α
📌 General Solutions for sin θ = sin α, cos θ = cos α, etc.
• sin θ = sin α ⇒ θ = nπ + (−1)ⁿα, n ∈ Z
• cos θ = cos α ⇒ θ = 2nπ ± α, n ∈ Z
• tan θ = tan α ⇒ θ = nπ + α, n ∈ Z
• cot θ = cot α ⇒ θ = nπ + α, n ∈ Z
• sec θ = sec α ⇒ θ = 2nπ ± α, n ∈ Z
• cosec θ = cosec α ⇒ θ = nπ + (−1)ⁿα, n ∈ Z
• cos θ = cos α ⇒ θ = 2nπ ± α, n ∈ Z
• tan θ = tan α ⇒ θ = nπ + α, n ∈ Z
• cot θ = cot α ⇒ θ = nπ + α, n ∈ Z
• sec θ = sec α ⇒ θ = 2nπ ± α, n ∈ Z
• cosec θ = cosec α ⇒ θ = nπ + (−1)ⁿα, n ∈ Z
JEE MATH APEX
3. Squared Trigonometric Equations
📌 Equations with sin²θ, cos²θ, tan²θ
• sin²θ = sin²α ⇒ θ = nπ ± α, n ∈ Z
• cos²θ = cos²α ⇒ θ = nπ ± α, n ∈ Z
• tan²θ = tan²α ⇒ θ = nπ ± α, n ∈ Z
• cot²θ = cot²α ⇒ θ = nπ ± α, n ∈ Z
• sec²θ = sec²α ⇒ θ = nπ ± α, n ∈ Z
• cosec²θ = cosec²α ⇒ θ = nπ ± α, n ∈ Z
• sin²θ = k (0 ≤ k ≤ 1) ⇒ θ = nπ ± sin⁻¹(√k), n ∈ Z
• cos²θ = cos²α ⇒ θ = nπ ± α, n ∈ Z
• tan²θ = tan²α ⇒ θ = nπ ± α, n ∈ Z
• cot²θ = cot²α ⇒ θ = nπ ± α, n ∈ Z
• sec²θ = sec²α ⇒ θ = nπ ± α, n ∈ Z
• cosec²θ = cosec²α ⇒ θ = nπ ± α, n ∈ Z
• sin²θ = k (0 ≤ k ≤ 1) ⇒ θ = nπ ± sin⁻¹(√k), n ∈ Z
JEE MATH APEX
4. Linear Equations in sin & cos
📌 a sin θ + b cos θ = c
• General form: a sin θ + b cos θ = c
• Solution exists if: |c| ≤ √(a² + b²)
• Put a = r cos α, b = r sin α where r = √(a² + b²)
• Then: r sin(θ + α) = c ⇒ sin(θ + α) = c/r
• Solution: θ = nπ + (−1)ⁿ sin⁻¹(c/r) − α, n ∈ Z
• Where α = tan⁻¹(b/a)
• Solution exists if: |c| ≤ √(a² + b²)
• Put a = r cos α, b = r sin α where r = √(a² + b²)
• Then: r sin(θ + α) = c ⇒ sin(θ + α) = c/r
• Solution: θ = nπ + (−1)ⁿ sin⁻¹(c/r) − α, n ∈ Z
• Where α = tan⁻¹(b/a)
JEE MATH APEX
5. Quadratic in Trigonometric Functions
📌 Quadratic Form Equations
• a sin²θ + b sin θ + c = 0 → Solve as quadratic in sin θ
• a cos²θ + b cos θ + c = 0 → Solve as quadratic in cos θ
• a tan²θ + b tan θ + c = 0 → Solve as quadratic in tan θ
• Method: Substitute t = sin θ (or cos θ or tan θ)
• Solve at² + bt + c = 0 → Then solve t = sin θ
• Check: −1 ≤ sin θ ≤ 1, −1 ≤ cos θ ≤ 1
• a cos²θ + b cos θ + c = 0 → Solve as quadratic in cos θ
• a tan²θ + b tan θ + c = 0 → Solve as quadratic in tan θ
• Method: Substitute t = sin θ (or cos θ or tan θ)
• Solve at² + bt + c = 0 → Then solve t = sin θ
• Check: −1 ≤ sin θ ≤ 1, −1 ≤ cos θ ≤ 1
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6. Equations with Multiple Angles
📌 sin nθ = sin α, cos nθ = cos α, etc.
• sin nθ = sin α ⇒ nθ = mπ + (−1)ᵐα → θ = [mπ + (−1)ᵐα]/n
• cos nθ = cos α ⇒ nθ = 2mπ ± α → θ = (2mπ ± α)/n
• tan nθ = tan α ⇒ nθ = mπ + α → θ = (mπ + α)/n
• For sin 2θ = sin α: 2θ = nπ + (−1)ⁿα
• For cos 3θ = cos α: 3θ = 2nπ ± α
• cos nθ = cos α ⇒ nθ = 2mπ ± α → θ = (2mπ ± α)/n
• tan nθ = tan α ⇒ nθ = mπ + α → θ = (mπ + α)/n
• For sin 2θ = sin α: 2θ = nπ + (−1)ⁿα
• For cos 3θ = cos α: 3θ = 2nπ ± α
JEE MATH APEX
7. Factorisation Method
📌 Sum to Product & Factorisation
• sin θ + sin α = 0 ⇒ 2 sin[(θ+α)/2] cos[(θ−α)/2] = 0
• sin θ − sin α = 0 ⇒ 2 cos[(θ+α)/2] sin[(θ−α)/2] = 0
• cos θ + cos α = 0 ⇒ 2 cos[(θ+α)/2] cos[(θ−α)/2] = 0
• cos θ − cos α = 0 ⇒ −2 sin[(θ+α)/2] sin[(θ−α)/2] = 0
• Then solve each factor separately.
• sin θ − sin α = 0 ⇒ 2 cos[(θ+α)/2] sin[(θ−α)/2] = 0
• cos θ + cos α = 0 ⇒ 2 cos[(θ+α)/2] cos[(θ−α)/2] = 0
• cos θ − cos α = 0 ⇒ −2 sin[(θ+α)/2] sin[(θ−α)/2] = 0
• Then solve each factor separately.
JEE MATH APEX
8. Equations Using Identities
📌 Transformation Using Identities
• sin θ + sin 3θ = 0 → Use sum to product
• cos θ + cos 2θ + cos 3θ = 0 → Group and factorise
• sin²θ + cos θ − 1 = 0 → Use sin²θ = 1 − cos²θ
• tan θ + cot θ = 2 → Convert to sin θ, cos θ
• sin θ + cos θ = 1 → Square both sides (check extraneous roots)
• cos θ + cos 2θ + cos 3θ = 0 → Group and factorise
• sin²θ + cos θ − 1 = 0 → Use sin²θ = 1 − cos²θ
• tan θ + cot θ = 2 → Convert to sin θ, cos θ
• sin θ + cos θ = 1 → Square both sides (check extraneous roots)
JEE MATH APEX
9. Homogeneous Equations
📌 Equations Homogeneous in sin θ and cos θ
• Form: a sin²θ + b sin θ cos θ + c cos²θ = 0
• Divide by cos²θ: a tan²θ + b tan θ + c = 0
• Form: a sin θ + b cos θ = 0
• Solution: tan θ = −b/a ⇒ θ = nπ + tan⁻¹(−b/a)
• Check if cos θ = 0 is also a solution
• Divide by cos²θ: a tan²θ + b tan θ + c = 0
• Form: a sin θ + b cos θ = 0
• Solution: tan θ = −b/a ⇒ θ = nπ + tan⁻¹(−b/a)
• Check if cos θ = 0 is also a solution
JEE MATH APEX
10. Conditional Equations (Triangle)
📌 If A + B + C = π (Triangle Conditions)
• tan A + tan B + tan C = tan A · tan B · tan C
• sin A + sin B + sin C = 4 cos(A/2) cos(B/2) cos(C/2)
• cos A + cos B + cos C = 1 + 4 sin(A/2) sin(B/2) sin(C/2)
• sin 2A + sin 2B + sin 2C = 4 sin A sin B sin C
• cos 2A + cos 2B + cos 2C = −1 − 4 cos A cos B cos C
• tan A/2 · tan B/2 + tan B/2 · tan C/2 + tan C/2 · tan A/2 = 1
• sin A + sin B + sin C = 4 cos(A/2) cos(B/2) cos(C/2)
• cos A + cos B + cos C = 1 + 4 sin(A/2) sin(B/2) sin(C/2)
• sin 2A + sin 2B + sin 2C = 4 sin A sin B sin C
• cos 2A + cos 2B + cos 2C = −1 − 4 cos A cos B cos C
• tan A/2 · tan B/2 + tan B/2 · tan C/2 + tan C/2 · tan A/2 = 1
JEE MATH APEX
11. Range & Domain Restrictions
📌 Checking Valid Solutions
• −1 ≤ sin θ ≤ 1 (Solutions outside this range are invalid)
• −1 ≤ cos θ ≤ 1
• tan θ, cot θ ∈ R (all real numbers)
• sec θ ≥ 1 or sec θ ≤ −1
• cosec θ ≥ 1 or cosec θ ≤ −1
• Always verify solutions in original equation (squaring may introduce extraneous roots)
• −1 ≤ cos θ ≤ 1
• tan θ, cot θ ∈ R (all real numbers)
• sec θ ≥ 1 or sec θ ≤ −1
• cosec θ ≥ 1 or cosec θ ≤ −1
• Always verify solutions in original equation (squaring may introduce extraneous roots)
JEE MATH APEX
12. Special Equations & Substitutions
📌 Advanced Techniques for JEE Advanced
• sin⁴θ + cos⁴θ = 1 → Use (sin²θ + cos²θ)² − 2sin²θcos²θ
• sin⁶θ + cos⁶θ = 1 − 3sin²θcos²θ
• sin θ · sin 2θ · sin 4θ = 0 → Each factor = 0
• Substitution: sin θ = t or tan(θ/2) = t
• sin θ + cos θ = t → sin θ cos θ = (t² − 1)/2
• |sin θ| + |cos θ| = 1 → θ = nπ/2, n ∈ Z
• sin⁶θ + cos⁶θ = 1 − 3sin²θcos²θ
• sin θ · sin 2θ · sin 4θ = 0 → Each factor = 0
• Substitution: sin θ = t or tan(θ/2) = t
• sin θ + cos θ = t → sin θ cos θ = (t² − 1)/2
• |sin θ| + |cos θ| = 1 → θ = nπ/2, n ∈ Z
JEE MATH APEX
13. Solutions in Given Interval
📌 Finding Solutions in [0, 2π] or [0, π]
• Find general solution first
• Substitute different integer values of n
• Collect only those values within the given interval
• Count number of solutions: sin θ = k (|k| < 1) has 2 solutions in [0, 2π]
• cos θ = k (|k| < 1) has 2 solutions in [0, 2π]
• tan θ = k has 2 solutions in [0, 2π]
• sin θ = 0 has 3 solutions in [0, 2π]: 0, π, 2π
• Substitute different integer values of n
• Collect only those values within the given interval
• Count number of solutions: sin θ = k (|k| < 1) has 2 solutions in [0, 2π]
• cos θ = k (|k| < 1) has 2 solutions in [0, 2π]
• tan θ = k has 2 solutions in [0, 2π]
• sin θ = 0 has 3 solutions in [0, 2π]: 0, π, 2π
✅ TRIGONOMETRIC EQUATIONS ✅
— JEE MATH APEX —
— JEE MATH APEX —
JEE MATH APEX 
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