Question

How do you solve: `sin^2(2x)+2cos^2(x) <2` ? Thank you

Answer 1

There are two families of solutions, with `n` being any integer:

`2n\pi+{\pi}/4\lt x\lt 2n\pi+{3\pi}/4`

`2n\pi+{5\pi}/4\lt x\lt 2n\pi+{7\pi}/4`

Explanation:

The best approach I can see is to convert the expression to one involving a single trigonometric function. To start we can use the double angle formula for sine:

`\sin^2(2x)=(2\sin(x)\cos(x))^2=4\sin^2(x)\cos^2(x)`

So then

`\sin^2(2x)+2cos^2(2x)=2\sin^2(x)\cos^2(x)+2cos^2(x)<2`

Then put `\sin^2(x)=1-\cos^2(x)`:

`6\cos^2(x)-4\cos^4(x)<2`

`4\cos^4(x)-6\cos^2(x)+2>0`

`2(2\cos^2(x)-1)(cos^2(x)-1)>0`

You must have `\cos(x)` between `-1` and `+1`, and the boundary values `\cos(x)=\pm1` would make the left side exactly zero. So `cos^2(x)-1<0` and we need:

`2cos^2(x)-1<0`

`-{\sqrt{2}}/2<\cos(x)<{\sqrt{2}}/2`

Therefore, using the fact that multiples of `{\pi}/4` give cosines equal to `\pm {\sqrt{2}}/2` there are two families of solutions:

`2n\pi+{\pi}/4\lt x\lt 2n\pi+{3\pi}/4`

`2n\pi+{5\pi}/4\lt x\lt 2n\pi+{7\pi}/4`

Answer 2

`((pi)/4, (3pi)/4); ((5pi)/4, (7pi)/4)`

Explanation:

`sin^2 (2x) + 2cos^2 x - 2 < 0`.
`sin^2 2x - 2(1 - cos ^2 x) < 0`
`sin^2 2x - 2sin^2 x < 0`
`4sin^2 x.cos^2 x - 2sin^2 x < 0`
`F(x) = 2sin^2 x(2cos^2 x - 1) < 0`
Call `f(x) = sin^2 x`, and `g(x) = (2cos^2 x - 1)`
Set up a sign chart for f(x) and g(x), when x varies from 0 to `2pi`.
The sign of F(x) is the resulting sign of the product f(x).g(x).
`f(x) = sin^2 x` is always positive regardless of x.
The sign of F(x) is the sign of g(x)
On the unit circle,
`g(x) = (2cos^2 x - 1) < 0` when x varies inside the intervals:
`(pi/4, (3pi)/4) and ((5pi)/4, (7pi)/4)`,
Inside these intervals --> `2cos^2 x < 1` --> `cos^2 x < 1/2`
Therefor, these intervals are the solution set for F(x) < 0.
Check with `x = pi/6 = 30^@`
`cos 30 = sqrt3/2` --> `cos^2 30 = 3/4` -->
`2cos^2 30 = 6/4 > 1`. Proved.