Trigonometry Trigonometric Equations 2 — Questions and Answers
Question 1: Solve sin²(x) = 1 for x in [0, 2π).
- x = 0, π
- x = π/2, 3π/2 (Correct answer)
- x = π/4, 3π/4, 5π/4, 7π/4
- x = π/2 only
Correct answer: x = π/2, 3π/2
sin²(x) = 1 means sin(x) = ±1, which occurs at x = π/2 and x = 3π/2 on [0, 2π).
Question 2: Solve 2cos²(x) − cos(x) − 1 = 0 for x in [0, 2π).
- x = π/3, π, 5π/3
- x = π/2, π, 3π/2
- x = π/6, π/2, 7π/6
- x = 0, 2π/3, 4π/3 (Correct answer)
Correct answer: x = 0, 2π/3, 4π/3
Factoring gives (2cos(x)+1)(cos(x)−1)=0, so cos(x)=−1/2 (x=2π/3,4π/3) or cos(x)=1 (x=0).
Question 3: Solve sin(x)cos(x) = 1/2 for x in [0, 2π).
- x = π/4, 3π/4
- x = π/4, 5π/4 (Correct answer)
- x = π/6, 5π/6
- x = π/3, 4π/3
Correct answer: x = π/4, 5π/4
Using the identity sin(x)cos(x) = (1/2)sin(2x), the equation becomes sin(2x)=1, giving 2x=π/2+2nπ, so x=π/4, 5π/4.
Question 4: What substitution simplifies solving 2sin²(x) + 3sin(x) − 2 = 0?
- Let u = cos(x)
- Let u = sin(x) (Correct answer)
- Let u = tan(x)
- Let u = sin(2x)
Correct answer: Let u = sin(x)
Substituting u = sin(x) converts the equation to the quadratic 2u² + 3u − 2 = 0, which can be factored easily.
Question 5: Solve cos(2x) = cos(x) for x in [0, 2π).
- x = 0, 2π/3, 4π/3 (Correct answer)
- x = π/3, π, 5π/3
- x = π/4, π/2, 7π/4
- x = 0, π/3, 5π/3
Correct answer: x = 0, 2π/3, 4π/3
Replacing cos(2x) with 2cos²(x)−1 and factoring gives (2cos(x)+1)(cos(x)−1)=0, yielding x=0, 2π/3, 4π/3.
Question 6: How many solutions does sin(x) = 0.8 have in the interval [0, 2π)?
- 0
- 1
- 2 (Correct answer)
- 4
Correct answer: 2
Since 0.8 is in (0,1), sine equals 0.8 once in Q1 and once in Q2, giving exactly 2 solutions per period.
Question 7: Solve tan²(x) − 3 = 0 for x in [0, 2π).
- x = π/6, 5π/6, 7π/6, 11π/6
- x = π/3, 2π/3, 4π/3, 5π/3 (Correct answer)
- x = π/4, 3π/4, 5π/4, 7π/4
- x = π/3, π, 5π/3
Correct answer: x = π/3, 2π/3, 4π/3, 5π/3
tan²(x)=3 gives tan(x)=±√3; tan=√3 at π/3,4π/3 and tan=−√3 at 2π/3,5π/3.
Trigonometry Trigonometric Equations 2
Solve sin²(x) = 1 for x in [0, 2π).