PIEAS Entry Test Mathematics Trigonometric Equations — Set 3

PIEAS Entry Test Mathematics Trigonometric Equations — Set 3

1. Question 1

   Q1. The solution of sin(x) = sin(π/3) is

   • A) nπ + (-1)^n * π/3
   • B) 2nπ + π/3
   • C) nπ + π/3
   • D) 2nπ - π/3

   Answer: nπ + (-1)^n * π/3

   Explanation: The general solution for sin(x) = sin(a) is x = nπ + (-1)^n * a, hence x = nπ + (-1)^n * π/3.

2. Question 2

   Q2. The general solution of cos(2x) = cos(π/4) is

   • A) nπ ± π/8
   • B) 2nπ ± π/8
   • C) nπ + π/8
   • D) 2nπ + π/8

   Answer: nπ ± π/8

   Explanation: For cos(2x) = cos(π/4), 2x = 2nπ ± π/4, hence x = nπ ± π/8.

3. Question 3

   Q3. If 2sin(x)cos(x) = 1, then x =

   • A) nπ + (-1)^n * π/4
   • B) nπ/2 + (-1)^n * π/4
   • C) nπ + π/4
   • D) nπ/2 - (-1)^n * π/4

   Answer: nπ/2 + (-1)^n * π/4

   Explanation: 2sin(x)cos(x) = sin(2x) = 1, hence 2x = nπ + (-1)^n * π/2, so x = nπ/2 + (-1)^n * π/4.

4. Question 4

   Q4. The solution of sin(3x) + sin(x) = 0 is

   • A) nπ/2
   • B) nπ
   • C) 2nπ
   • D) nπ/4

   Answer: nπ

   Explanation: Using sum-to-product, 2sin(2x)cos(x) = 0, hence sin(2x) = 0 or cos(x) = 0, giving x = nπ/2 or (2n+1)π/2, simplifying to nπ.

5. Question 5

   Q5. The general solution of cos(x) + cos(3x) = 0 is

   • A) nπ/2 + π/4
   • B) nπ + π/4
   • C) nπ ± π/4
   • D) nπ/2 ± π/4

   Answer: nπ/2 ± π/4

   Explanation: Using sum-to-product, 2cos(2x)cos(x) = 0, hence cos(2x) = 0 or cos(x) = 0, giving x = nπ/2 ± π/4 or (2n+1)π/2.

6. Question 6

   Q6. The solution of 3tan²(x) - 1 = 0 is

   • A) nπ ± π/6
   • B) nπ ± π/3
   • C) 2nπ ± π/6
   • D) 2nπ ± π/3

   Answer: nπ ± π/6

   Explanation: tan²(x) = 1/3, hence tan(x) = ±1/√3, so x = nπ ± π/6.

7. Question 7

   Q7. If sin(2x) + sin(4x) = 0, then x =

   • A) nπ/3
   • B) nπ/2
   • C) nπ
   • D) nπ/6

   Answer: nπ/3

   Explanation: Using sum-to-product, 2sin(3x)cos(x) = 0, hence sin(3x) = 0, giving 3x = nπ, so x = nπ/3.

8. Question 8

   Q8. The solution of tan(x) + tan(2x) + tan(3x) = 0 is

   • A) nπ/3
   • B) nπ/2
   • C) nπ
   • D) nπ/6

   Answer: nπ/3

   Explanation: Using the tangent sum identity, the equation simplifies to tan(6x) = 0, hence 6x = nπ, so x = nπ/6, but upon detailed simplification, it results in nπ/3.

9. Question 9

   Q9. The general solution of 2sin(3x) + 1 = 0 is

   • A) nπ/3 + (-1)^(n+1) * π/18
   • B) nπ + (-1)^(n+1) * π/6
   • C) nπ/3 + (-1)^n * π/18
   • D) nπ + (-1)^n * π/6

   Answer: nπ/3 + (-1)^(n+1) * π/18

   Explanation: sin(3x) = -1/2, hence 3x = nπ + (-1)^(n+1) * π/6, so x = nπ/3 + (-1)^(n+1) * π/18.

10. Question 10

    Q10. The solution of cos(3x) = cos(2x) is

    • A) 2nπ/5
    • B) 2nπ
    • C) nπ
    • D) 2nπ or 2nπ/5

    Answer: 2nπ or 2nπ/5

    Explanation: For cos(3x) = cos(2x), 3x = 2nπ ± 2x, giving x = 2nπ or 2nπ/5.

11. Question 11

    Q11. Solve: sin(2x) = sin(x)

    • A) 2nπ or nπ + (-1)^n π/3
    • B) nπ or 2nπ ± π/3
    • C) 2nπ/3 or nπ
    • D) nπ or 2nπ/3

    Answer: 2nπ or nπ + (-1)^n π/3

    Explanation: Using the identity sin(a) = sin(b), we get 2x = nπ + (-1)^n x, simplifying to the given solutions.

12. Question 12

    Q12. The general solution of tan(2x) = cot(x) is

    • A) nπ/2 + π/6
    • B) nπ/2 - π/6
    • C) nπ + π/6
    • D) nπ - π/6

    Answer: nπ/2 + π/6

    Explanation: tan(2x) = 1/tan(x) gives tan(2x)tan(x) = 1, simplifying to 2x = nπ + π/2 - x, hence 3x = nπ + π/2.

13. Question 13

    Q13. If cos(2x) + 2cos(x) + 1 = 0, then x =

    • A) (2n + 1)π
    • B) 2nπ
    • C) (2n + 1)π/2
    • D) nπ

    Answer: (2n + 1)π

    Explanation: The equation simplifies to (cos(x) + 1)² = 0, so cos(x) = -1, giving x = (2n + 1)π.

14. Question 14

    Q14. The solution of 3tan²(x) = 1 is

    • A) nπ ± π/6
    • B) nπ ± π/3
    • C) 2nπ ± π/6
    • D) 2nπ ± π/3

    Answer: nπ ± π/6

    Explanation: tan²(x) = 1/3 gives tan(x) = ±1/√3, so x = nπ ± π/6.

15. Question 15

    Q15. If 2sin²(x) + sin²(2x) = 2, then x =

    • A) nπ ± π/4
    • B) nπ ± π/2
    • C) nπ/2 ± π/4
    • D) nπ ± π/3

    Answer: nπ ± π/4

    Explanation: Simplifying gives 2sin²(x) + 4sin²(x)cos²(x) = 2, or 2sin²(x)(1 + 2cos²(x)) = 2, simplifying to sin²(x) = 1/2.

16. Question 16

    Q16. Solve: tan(x) + tan(2x) + tan(3x) = 0

    • A) nπ/3 or nπ ± π/6
    • B) nπ/3
    • C) nπ ± π/6
    • D) nπ/6

    Answer: nπ/3

    Explanation: Using the tangent sum identity, we simplify to tan(3x)(1 - tan(x)tan(2x)) + tan(3x) = 0, giving tan(3x) = 0, hence x = nπ/3.

17. Question 17

    Q17. If sin(x) + cos(x) = 1, then x =

    • A) 2nπ or 2nπ + π/2
    • B) 2nπ
    • C) 2nπ + π/2
    • D) nπ

    Answer: 2nπ or 2nπ + π/2

    Explanation: Squaring both sides and simplifying gives sin(2x) = 0, so 2x = 2nπ or 2x = 2nπ + π, giving x = nπ or nπ + π/2, but checking gives 2nπ or 2nπ + π/2.

18. Question 18

    Q18. If tan(2x) = tan(x), then x =

    • A) nπ
    • B) nπ/2
    • C) nπ + π/2
    • D) nπ - π/2

    Answer: nπ

    Explanation: tan(2x) - tan(x) = 0 gives sin(x)/cos(2x) = 0, so sin(x) = 0, hence x = nπ.

19. Question 19

    Q19. If 2sin(x) + 1 = 0, then x = ?

    • A) 7π / 6
    • B) 11π / 6
    • C) Both A and B
    • D) π / 6

    Answer: Both A and B

    Explanation: General solution for sin(x) = -1/2 is x = 7π/6 + 2kπ or 11π/6 + 2kπ, k ∈ Z.

20. Question 20

    Q20. Solve: tan(x) = √3

    • A) π / 3
    • B) 4π / 3
    • C) Both A and B
    • D) 2π / 3

    Answer: Both A and B

    Explanation: tan(x) = √3 for x = π/3 + kπ, k ∈ Z, so x = π/3 or 4π/3.