Question

on interval 0 ≤ x < 2π, where are the x - intercepts of y = cos(2x)?
□ $\frac{pi}{2}$ and $\frac{3pi}{2}$
□ 0, π, and 2π
□ $\frac{pi}{2}$, π and $\frac{3pi}{2}$
□ $\frac{pi}{4}$, $\frac{3pi}{4}$, $\frac{5pi}{4}$ and $\frac{7pi}{4}$

Explanation:

Step1: Set y = 0

Set $\cos(2x)=0$.

Step2: Use cosine - zero property

We know that $\cos\theta = 0$ when $\theta=(2n + 1)\frac{\pi}{2}$, where $n$ is an integer. So, $2x=(2n + 1)\frac{\pi}{2}$.

Step3: Solve for x

$x=(2n + 1)\frac{\pi}{4}$.

Step4: Find values of n for the given interval

For $0\leq x<2\pi$, when $n = 0$, $x=\frac{\pi}{4}$; when $n = 1$, $x=\frac{3\pi}{4}$; when $n = 2$, $x=\frac{5\pi}{4}$; when $n = 3$, $x=\frac{7\pi}{4}$.

Answer:

$\frac{\pi}{4},\frac{3\pi}{4},\frac{5\pi}{4},\frac{7\pi}{4}$