Question

How do you solve `sin^2x < 1/2`?

Solution:

Answer 1

Given: `sin^2(x)<1/2`

When we take the square root of both sides, we obtain two equations:

`sin(x)> -1/sqrt2` and `sin(x) < 1/sqrt2`

Take the inverse sine of both sides of both equations:

`x> sin^-1(-1/sqrt2)` and `x < sin^-1(1/sqrt2)`

We know that the primary values are `-pi/4 = sin^-1(-1/sqrt2)` and `pi/4 = sin^-1(1/sqrt2)`:

`x> -pi/4` and `x < pi/4`

This can be written as a single interval:

`-pi/4 < x < pi/4`

We know that it is periodic with integer multiples of `pi`:

`-pi/4 + npi < x < pi/4 + npi, n in ZZ`

Answer 2

`(-pi/4, pi/4), and ((3pi)/4 , (5pi)/4)`

Explanation:

First, find the 4 end-points, by solving `f(x) = sin^2 x = 1/2`
a. `sin x = sqrt2/2` --> `x = pi/4 and x = (3pi)/4`
b. `sin x = - sqrt2/2` --> `x = - pi/4, and x = (5pi)/4`
Now, solve the trig inequality by using the unit circle as proof.
`f(x) = sin^2 x < 1/2`
`Isin xI < sqrt2/2`
On the unit circle, `Isin xI < sqrt2/2`, when the arc x varies inside the
2 open intervals, that are the answers:
`(-pi/4, pi/4) and ((3pi)/4, (5pi)/4)`