Since trigonometrical functions are periodic functions, solutions of trigonometric equations can be generalized with the help of the periodicity of the trigonometrical functions. The solution consisting of all possible solutions of a trigonometric equation is called its general solution.
Fundamental Formulas for Solving Trigonometric Equations
sin θ = 0 ⟹ θ = nπ
cos θ = 0 ⟹ θ = (2n + 1)π/2
tan θ = 0 ⟹ θ = nπ
sin θ = sin α ⟹ θ = nπ + (–1)ⁿα, where α ∈ [–π/2, π/2]
cos θ = cos α ⟹ θ = 2nπ ± α, where α ∈ [0, π]
tan θ = tan α ⟹ θ = nπ + α, where α ∈ (–π/2, π/2)
sin²θ = sin²α, cos²θ = cos²α, tan²θ = tan²α ⟹ θ = nπ ± α
sin θ = 1 ⟹ θ = (4n + 1)π/2
cos θ = 1 ⟹ θ = 2nπ
cos θ = –1 ⟹ θ = (2n + 1)π
sin θ = ± sin α and cos θ = ± cos α ⟹ θ = 2nπ + α
Important Points to Remember
- While solving a trigonometric equation, squaring the equation at any step should be avoided as far as possible. If squaring is necessary, check the solution for extraneous roots.
- Never cancel terms containing unknown terms on the two sides, which are in product. It may cause loss of genuine solution.
- The answer should not contain such values of angles which make any of the terms undefined or infinite.
- Domain should not be changed. If it changes, necessary corrections must be made.
- Check that denominator is not zero at any stage while solving equations.
- At times you may find that your answers differ from those in the package in their notations. This may be due to the different methods of solving the same problem. Whenever you come across such situation, you must check their identity.
- While solving trigonometric equations, you may get same set of solutions repeated in your answer. It is necessary for you to exclude those repetitions.
For example, in nπ + π/2, kπ/5 + π/10 (n, k ∈ I), the set nπ + π/2 forms a part of the second set of solutions (you can check by putting k = 5m + 2 (m ∈ I). Hence the correct answer would be kπ/5 + π/10, k ∈ I.
- Sometimes the two solution sets consist partly of common values. In all such cases the common part must be presented only once.
Illustration 1: For the equation sin²θ – cos θ = 1/4, the values of θ in the interval 0 ≤ θ ≤ 2π, are
(A) π/6, 5π/6
(B) 7π/6, 11π/6
(C) π/3, 2π/3
(D) π/3, 5π/3
Solution: (D)
The given equation can be written as 1 – cos²θ – cos θ = 1/4
Illustration 2:
The general solutions of the equation sin 6θ = sin 4θ – sin 2θ are given by
(A) 2nπ ± π/6, n ∈ I
(B) 2nπ ± π/4, n ∈ I
(C) nπ ± π/4, m, m, n ∈ I
(D) (2l + 1)π/6, (2m + 1)π/4, nπ, l, m, n ∈ I
Solution: (D)
sin 6θ = sin 4θ – sin 2θ ⟹ 2 sin 3θ cos 3θ = 2 cos 3θ. sin θ
⟹ 2 sin 3θ cos 3θ – 2 cos 3θ. sin θ = 0
⟹ 2 cos 3θ (sin 3θ – sin θ) = 0 ⟹ 4 cos 3θ cos 2θ sin θ = 0
Illustration 3:
The most general solution of the equation cos θ + √3 sin θ = √2 is
(A) nπ + (-1)ⁿπ/4, n ∈ I
(B) nπ + π/4, n ∈ I
(C) nπ + (-1)ⁿπ/6, n ∈ I
(D) none of these
Solution: (D)
Dividing the equation by √(1² + (√3)²) = √4 = 2
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Exercise Problems
Exercise 1:
(i) The general solution of the equation 7cos²θ + 3 sin²θ = 4 is given by
(A) 2nπ ± π/3, n ∈ I
(B) 2nπ ± π/3, n ∈ I
(C) 2nπ ± π/3, 2mπ ± π/3, m,n ∈ I
(D) nπ ± π/3, n ∈ I
(ii) The general solution of the equation tan²θ + cot²θ = 2 is given by
(A) 2nπ ± π/4, n ∈ I
(B) nπ ± π/4, n ∈ I
(C) nπ + π/4, n ∈ I
(D) none of these
(iii) The general solution of the equation 3tan²θ – 2sin θ = 0 is
(A) nπ, mπ + (-1)ⁿπ/6, m, n ∈ I
(B) nπ + (-1)ⁿπ/6, n ∈ I
(C) nπ, n ∈ I
(D) nπ + (-1)ⁿπ/3, n ∈ I
Exercise 1 Answers:
(i) C
(ii) B
(iii) A
Formulae and Concepts at a Glance
- While solving simultaneous equations, after getting the principle solution add 2nπ to get the general solution.
- While handling problems of trigonometric inequalities, graphical method of solution is very helpful. After getting the principle range add 2nπ to both extremes to get the general solution.
- While solving a trigonometric equation, squaring the equation at any step should be avoided as far as possible. If squaring is necessary, check the solution for extraneous roots.
- Never cancel terms containing unknown terms on the two sides, which are in product. It may cause loss of genuine solution.
- Check that denominator is not zero at any stage while solving equations.
- While solving trigonometric equations, you may get same set of solution repeated in your answer it is necessary for you to exclude those repetitions.
- Some times the two solution set consist partly of common values. In all such cases the common part must be presented only once.
Solved Examples
1. The general solution of the trigonometric equation sin x + cos x = 1, for n = 0, 1, 2,… is given by
(A) x = 2nπ
(B) x = 2nπ + π/2
(C) x = nπ + (-1)ⁿ(π/4) - π/4
(D) none of these
Solution: We write the given equation as
(1/√2)sin x + (1/√2)cos x = 1/√2
or cos(π/4) sin x + sin (π/4) cos x = sin (π/4) or sin (x + π/4) = sin (π/4)
⟹ x + π/4 = nπ + (-1)ⁿ(π/4)
or x = nπ + (-1)ⁿ(π/4) - (π/4) when n = 0, 1, 2,…
2. The number of all possible triplets (a₁, a₂, a₃) such that a₁ + a₂ cos 2x + a₃ sin² x = 0 for all x is
(A) zero
(B) one
(C) three
(D) infinite
Solution: (D) We write the given relation as
a₁ + a₂cos2x + a₃(1 - cos2x)/2 = 0
Choosing a₃ = k, k∈ℝ, we get, a₁ = -k/2, a₂ = k/2.
Hence the solution set is (-k/2, k/2, k), where k is any real number. Thus the number of solutions is infinite.
3. The solution set of (2cos x - 1) (3 + 2cos x) = 0 in the interval 0 ≤ x ≤ 2π is
(A) {π/3}
(B) {π/3, 5π/3}
(C) {π/3, 5π/3, cos⁻¹(-3/2)}
(D) none of these
Solution: (B) We have 2cos x - 1 = 0 or 3 + 2cos x = 0
But 3 + 2cos x = 0 gives cos x = -3/2 which is not possible.
From 2cos x - 1 = 0, we get cos x = 1/2 whence x = π/3, 2π - π/3 i.e. 5π/3 in the interval 0 ≤ x ≤ 2π.
Thus the solution set in the given interval is {π/3, 5π/3}.
Assignment Problems
1. The value of a for which the equation 4cosec²(π(a + x)) + a² - 4a = 0 has a real solution, is
(A) a = 1 (B) a = 2 (C) a = 10 (D) none of these
2. The set of all values of x in the interval [0, π] for which 2sin²x - 3sinx + 1 ≥ 0 contains
(A) [0, π/6] (B) [0, π/3] (C) [2π/3, π] (D) [5π/6, π/2]
3. The number of solutions of 2cos(x/2) = 3ˣ + 3⁻ˣ, x ∈ [0,2π] is
(A) 0 (B) 1 (C) 2 (D) infinite
4. If the equation x - cos θ = sin θ has a real solution in θ, then
(A) x ≥ √2 (B) x ≤ √2 (C) |x| ≥ √2 (D) |x| ≤ √2
5. If -π/2 < x < π/2 and sin²x + cos²x = 81/81 = 30, then x is equal to
(A) π/3 or π/6 (B) 2π/3 or 5π/6 (C) π/6 or 5π/6 (D) none of these
6. If 1/(sec θ - tan θ) = 2, then θ lies in
(A) the first quadrant (B) the second quadrant (C) the third quadrant (D) the fourth quadrant
7. The number of solutions of the equation 3 cos θ + 4 sin θ = 6 is
(A) 0 (B) 1 (C) 2 (D) 3
8. The number of values of x lying in the interval (-π,π) which satisfy the equation (1 + |cos x| + cos²x + |cos³x| + ...) = 8/3 is
(A) 1 (B) 2 (C) 3 (D) 4
9. If sin⁴x + cos⁴x = a has a real solution, then
(A) 1/2 ≤ a ≤ 1 (B) 1/2 ≤ a ≤ 2 (C) 1/2 ≤ a ≤ 3/2 (D) none of these
10. The general value of θ if log₅ tan θ = log₅ 4.log₄ (3 sin θ) is
(A) 2nπ + cos⁻¹ 1/3 (B) 2nπ - cos⁻¹ 1/3 (C) 2nπ (D) 0
11. The number of solutions of the equation sin³x cos x + sin²x cos²x + sin x cos³x = 1 in the interval [0, 2π] is
(A) 0 (B) 2 (C) 3 (D) infinite
12. If tan(θ/2) = sin(α/2)/cos(α/2), then sin θ + cos θ is equal to
(A) 0 (B) 1 (C) -1 (D) 1 or -1
13. If (1 + sin A)/cos A = (1 + sin B)/cos B, then the values of A and B lying between 0 and 90° are respectively
(A) 30° and 60° (B) 60° and 30° (C) 45° and 15° (D) none of these
14. Solution of the equation (√2 + 1)^(5cos²θ + 2cosθ - 1) = 0, -π ≤ θ ≤ π is
(A) π/4 (B) ±π/3, cos⁻¹(3/5) (C) ±π/3, ±cos⁻¹(3/5) (D) none of these
15. The number of solutions of the equation cos θ cos 2θ cos 3θ = 1/4; 0 ≤ θ ≤ π is
(A) 1 (B) 3 (C) 5 (D) 6
Answers to Assignment Problems
| Problem | Answer | Problem | Answer | Problem | Answer |
|---|---|---|---|---|---|
| 1 | B | 6 | A | 11 | A |
| 2 | A | 7 | A | 12 | D |
| 3 | B | 8 | D | 13 | C |
| 4 | D | 9 | A | 14 | C |
| 5 | B | 10 | A | 15 | D |