Answer:
The zeroes of the f(x) = ± π/4 , ±3π/4 , ±5π/4 , ±7π/4 ⇒ answer (d)
Step-by-step explanation:
* The domain of the function f(x) = 1/2 cos(2x) is [-2π , 2π]
* That means -2π ≤ x ≤ 2π
∴ The domain of 2x is ⇒ -4π ≤ 2x ≤ 4π
* lets solve for 2x and then find the values of x
∵ f(x) = 1/2 cos(2x)
∴ The zeroes of the function means f(x) = 0
∴ 1/2 cos(2x) = 0 ⇒ multiply both sides by 2
∴ cos(2x) = 0
* In the four quadrant the values of cosФ = 0 at the y-axis with angles
π/2 , 3π/2 (anti-clockwise) , -π/2 , -3π/2 (clockwise) for 1 cycle,
but we have 2 cycles for 2x, so we add 2π for every angle of
2x, then we have other 4 angles 5π/2 , 7π/2 (anti-clockwise),
-5π/2 , -7π/2 (clockwise)
∴ 2x = ± π/2 , ±3π/2 , ±5π/2 , ±7π/2
* To find the value of x divide all by 2
∴ x = ± π/4 , ±3π/4 , ±5π/4 , ±7π/4
* The zeroes of the f(x) = ± π/4 , ±3π/4 , ±5π/4 , ±7π/4
