Multiple Choice Questions
Algebra: Trigonometry and Trigonometric Equations
Topic: Trigonometric Equations
Grade: 11
Question 1:
Solve the equation 2sin^2(x) – 3sin(x) + 1 = 0 for 0 ≤ x ≤ 2Ï€.
A) x = 0, x = π
B) x = π/3, x = 5π/3
C) x = π/6, x = 7π/6
D) x = π/4, x = 3π/4
Answer: C) x = π/6, x = 7π/6
Explanation: To solve this equation, we can factorize it as (2sin(x) – 1)(sin(x) – 1) = 0. Setting each factor equal to zero, we get sin(x) = 1/2 and sin(x) = 1. The solutions for sin(x) = 1/2 are x = Ï€/6 and x = 7Ï€/6, and the solution for sin(x) = 1 is x = Ï€/2. However, since the given range is 0 ≤ x ≤ 2Ï€, the only valid solutions are x = Ï€/6 and x = 7Ï€/6.
Example 1: For x = π/6, sin(x) = 1/2. This means that the angle whose sine is 1/2 is π/6 radians or 30 degrees. Similarly, for x = 7π/6, sin(x) = 1/2, which corresponds to an angle of 7π/6 radians or 210 degrees.
Example 2: If we substitute x = Ï€/2 into the equation, we get 2sin^2(Ï€/2) – 3sin(Ï€/2) + 1 = 0. Simplifying this expression, we get 2 – 3 + 1 = 0, which is not true. Therefore, x = Ï€/2 is not a valid solution for the given equation.
Question 2:
Find the general solution for the equation cos(2x) = -1.
A) x = π/4 + nπ/2, where n is an integer
B) x = π/2 + nπ, where n is an integer
C) x = π/2 + nπ/2, where n is an integer
D) x = π/4 + nπ, where n is an integer
Answer: D) x = π/4 + nπ, where n is an integer
Explanation: To find the general solution for the equation cos(2x) = -1, we need to solve for x. Taking the inverse cosine of both sides, we get 2x = π + 2nπ or 2x = 3π + 2nπ, where n is an integer. Simplifying these equations, we get x = π/2 + nπ/2 or x = 3π/2 + nπ/2. However, since we are looking for the general solution, we can combine these two equations as x = π/4 + nπ.
Example 1: For n = 0, x = π/4. Substituting this value into the equation, we get cos(2(π/4)) = -1, which is true.
Example 2: For n = 1, x = 5Ï€/4. Substituting this value into the equation, we get cos(2(5Ï€/4)) = -1, which is also true.