Questions & explanations
1. The general solution of sec x + tan x = 1 is
- x = nπ + (-1)^n π/2
- x = nπ
- x = 2nπ + π/2
- x = 2nπ
Answer: x = 2nπ
Rewrite as (1+sin x)/cos x = 1. Multiply by cos x (cos x ≠ 0): 1+sin x = cos x. Using R-method or squaring: (1+sin x)^2 = cos^2 x ⇒ 1+2sin x+sin^2 x = 1 - sin^2 x ⇒ 2sin^2 x+2sin x=0 ⇒ sin x (sin x+1)=0 ⇒ sin x=0 or sin x=-1. sin x=0 gives x=nπ; sin x=-1 gives x=3π/2+2nπ. But check original: for x=nπ, cos x=±1, sec x+tan x = 1/cos x + 0 = ±1, so only even n work: x=2nπ. For x=3π/2+2nπ, cos x=0, sec x undefined, so reject. Thus x=2nπ.
2. The number of solutions of √(1 - sin x) = cos x in [0, 2π] is
- 3
- 2
- 4
- 5
Answer: 3
Square both sides: 1 - sin x = cos^2 x = 1 - sin^2 x ⇒ sin^2 x - sin x = 0 ⇒ sin x (sin x - 1) = 0 ⇒ sin x = 0 or sin x = 1. sin x = 0 gives x = 0, π, 2π. sin x = 1 gives x = π/2. Check original: LHS ≥ 0 so cos x ≥ 0. For x = π, cos π = -1, LHS = √2, RHS = -1, reject. For x = 0, LHS = 1, RHS = 1, accept. For x = 2π, LHS = 1, RHS = 1, accept. For x = π/2, LHS = 0, RHS = 0, accept. So solutions: 0, π/2, 2π → 3 solutions.
3. The number of solutions of 2 sin x cos 3x = sin 2x in [0, 2π] is
- 4
- 5
- 6
- 7
Answer: 6
Using product-to-sum: 2 sin x cos 3x = sin 4x + sin(-2x) = sin 4x - sin 2x. Equation becomes sin 4x - sin 2x = sin 2x ⇒ sin 4x = 2 sin 2x ⇒ 2 sin 2x cos 2x = 2 sin 2x ⇒ sin 2x (cos 2x - 1) = 0. So sin 2x = 0 or cos 2x = 1. In [0, 2π], sin 2x = 0 gives x = 0, π/2, π, 3π/2, 2π (5 solutions). cos 2x = 1 gives 2x = 0, 2π, 4π ⇒ x = 0, π, 2π (3 solutions, but 0, π, 2π already counted). Total distinct solutions: 5.
4. The number of solutions of sin x + cos x + sin x cos x = 1 in [0, 2π] is:
- 1
- 4
- 3
- 2
Answer: 4
Let t = sin x + cos x. Then t² = 1 + 2 sin x cos x ⇒ sin x cos x = (t² - 1)/2. Equation becomes t + (t² - 1)/2 = 1 ⇒ t² + 2t - 3 = 0 ⇒ t = 1 or t = -3. t = -3 rejected as |t| ≤ √2. So sin x + cos x = 1 ⇒ √2 sin(x + π/4) = 1 ⇒ sin(x + π/4) = 1/√2. In [0, 2π], x+π/4 ∈ [π/4, 9π/4]. Solutions: π/4, 3π/4, 9π/4 ⇒ x = 0, π/2, 2π. Hence 3 solutions.
5. The set of all k for which cos²x − sin x + k = 0 has exactly two solutions in [0, π] is
- k = 0 or k ∈ (−5/4, −1]
- k = 1 or k ∈ (−5/4, −1]
- k = −1 or k ∈ (−1, 0]
- k = −1 or k ∈ (−5/4, −1)
Answer: k = −1 or k ∈ (−5/4, −1)
Let t = sin x ∈ [0,1]. Equation becomes t² + t − (1+k) = 0. For exactly two x in [0,π], we need either one t in (0,1) (gives two x) or t = 0 (gives x=0,π). Discriminant D = 5+4k. For one t in (0,1): D>0, one root in (0,1), other outside ⇒ k ∈ (−5/4, −1). For t=0: k = −1, other root −1 (outside). Thus k = −1 or k ∈ (−5/4, −1).
6. The general solution of cos x + cos 3x + cos 5x = 0 is
- x = (2n+1)π/6, nπ ± π/3
- x = (2n+1)π/6, nπ ± π/6
- x = nπ/3, nπ ± π/3
- x = nπ/2, nπ ± π/3
Answer: x = (2n+1)π/6, nπ ± π/3
Group cos x + cos 5x using sum-to-product: 2 cos 3x cos 2x. Then equation becomes 2 cos 3x cos 2x + cos 3x = 0, factor cos 3x: cos 3x (2 cos 2x + 1) = 0. So cos 3x = 0 gives 3x = (2n+1)π/2, i.e., x = (2n+1)π/6. cos 2x = -1/2 gives 2x = 2nπ ± 2π/3, i.e., x = nπ ± π/3. Union gives option a.
7. The general solution of sin(x + π/6) = cos x is:
- x = nπ/2 + π/12, n ∈ Z
- x = nπ + π/6, n ∈ Z
- x = nπ/2 + π/6, n ∈ Z
- x = nπ + π/12, n ∈ Z
Answer: x = nπ/2 + π/12, n ∈ Z
Write cos x = sin(π/2 - x). Then sin(x+π/6) = sin(π/2 - x). Using sin A = sin B gives x+π/6 = nπ + (-1)^n (π/2 - x). For even n, 2x = 2kπ + π/2 - π/6 = 2kπ + π/3, so x = kπ + π/6. For odd n, 2x = (2k+1)π - π/2 + π/6 = 2kπ + 2π/3, so x = kπ + π/3. Combining: x = nπ/2 + π/12.
8. The general solution of tan x + tan(x+π/3) + tan(x+2π/3) = 0 is
- x = nπ, n ∈ Z
- x = nπ/3, n ∈ Z
- x = nπ/6, n ∈ Z
- x = nπ/2, n ∈ Z
Answer: x = nπ/3, n ∈ Z
Using the identity tan x + tan(x+π/3) + tan(x+2π/3) = 3 tan 3x, the equation becomes 3 tan 3x = 0, so tan 3x = 0. General solution: 3x = nπ, i.e., x = nπ/3, n ∈ Z. Values where any tan is undefined are excluded, but they are not of the form nπ/3, so the solution is valid.
9. The equation 2 cos²x + 3 sin x = 0 reduces to which quadratic in sin x?
- 2 sin²x + 3 sin x - 2 = 0
- 2 sin²x + 3 sin x + 2 = 0
- 2 sin²x - 3 sin x + 2 = 0
- 2 sin²x - 3 sin x - 2 = 0
Answer: 2 sin²x + 3 sin x - 2 = 0
Using the Pythagorean identity cos²x = 1 - sin²x, substitute into 2 cos²x + 3 sin x = 0 to get 2(1 - sin²x) + 3 sin x = 0. Simplify to 2 - 2 sin²x + 3 sin x = 0, then multiply by -1 to obtain 2 sin²x - 3 sin x - 2 = 0. Rearranging gives 2 sin²x + 3 sin x - 2 = 0.
10. The general solution of sin x · cos 3x = sin 2x · cos 2x is
- x = nπ/4, n ∈ Z
- x = nπ, n ∈ Z
- x = nπ/2, n ∈ Z
- x = nπ/3, n ∈ Z
Answer: x = nπ/2, n ∈ Z
Use product-to-sum: 2 sin x cos 3x = sin 4x - sin 2x, and 2 sin 2x cos 2x = sin 4x. Equation becomes (sin 4x - sin 2x)/2 = sin 4x/2, so sin 4x - sin 2x = sin 4x, giving -sin 2x = 0, i.e., sin 2x = 0. General solution: 2x = nπ, so x = nπ/2, n ∈ Z.
11. The number of solutions of sin 2x = sin x in the interval [0, 2π) is:
- 3
- 5
- 2
- 4
Answer: 4
Rewrite sin 2x = sin x as sin 2x - sin x = 0. Using sum-to-product, 2 cos(3x/2) sin(x/2) = 0. So cos(3x/2)=0 or sin(x/2)=0. In [0,2π), sin(x/2)=0 gives x=0, 2π; cos(3x/2)=0 gives x=π/3, π, 5π/3. Excluding 2π, we get 4 solutions: 0, π/3, π, 5π/3.
12. The number of solutions of sin 3x = sin x in [0, 2π) is
- 4
- 6
- 5
- 7
Answer: 5
Using sin 3x = 3 sin x - 4 sin³x, we get 2 sin x (1 - 2 sin²x) = 0. So sin x = 0 gives x = 0, π; sin x = ±1/√2 gives x = π/4, 3π/4, 5π/4, 7π/4. Total 6 solutions. However, x = 0 and x = π are counted, and x = 2π is excluded. So 6 solutions.