Question

What are the `x`-intercepts of the graph of `f(x)=8 sin^2x-4` on the interval `[0, 2pi]`?

Answer 1

`x = pi/4, (3pi)/4, (5pi)/4, (7pi)/4` for `x in [0, 2pi]`

Explanation:

`f(x) = 8sin^2x-4`

The `x-`intercepts of the graph of `f(x)` occur where `f(x)=0`

Hence, where: `8sin^2x-4 = 0`

`8sin^2x = 4`

`sin^2x = 4/8 = 1/2`

`:. sinx = +-sqrt(1/2)`

`x = arcsin (+-1/sqrt2)`

`x = pi/4, (3pi)/4, (5pi)/4, (7pi)/4` for `x in [0, 2pi]`

Which generalises to `(2n-1)pi/4 forall n in ZZ`

This can be seen on the graph of `f(x)` below.

graph{8(sinx)^2-4 [-8.78, 9, -4.17, 4.71]}