Question

How to solve this inequality `|sin2x|>sqrt3/2` ?

`|sin2x|>sqrt3/2`

Answer 1

`... uu (-{5pi}/6,-{2pi}/3) uu (-pi/3,-pi/6) uu(pi/6,pi/3) uu ({2pi}/3,{5pi}/6) uu ({7pi}/6,{4pi}/3) uu ({5pi}/3,{11pi}/6) uu ...`

Explanation:

We can say more generally that we want to find the values of `x` which satisfy

`sqrt3/2 < sin2x` or `sin2x < sqrt3/2`.

The period of `sin(2x)` is `pi`, i.e. it repeats itself after an interval of `pi`. Mathematically, we write `sin(2x)=sin(2x+pi)` for any `x in RR`. A graph of `y=sin(2x)` is appended below.

graph{sin(2x) [-10, 10, -5, 5]}

Therefore, we will limit `x in [0,pi)` first.

We first attempt to solve the first inequality.

`sqrt3/2 < sin2x`

Since `sin` is positive in the first and second quadrant, and we know that the basic angle is `sin^{-1}(sqrt3/2)`, we get

`sin^{-1}(sqrt3/2) < 2x < pi-sin^{-1}(sqrt3/2)`

`pi/3 < 2x < pi-pi/3`

`pi/3 < 2x < {2pi}/3`

`pi/6 < x < pi/3`

For the second inequality,

`sin2x < sqrt3/2`

Since `sin` is negative in the third and forth quadrant, we get

`pi + sin^{-1}(sqrt3/2) < 2x < 2pi-sin^{-1}(sqrt3/2)`

`pi + pi/3 < 2x < 2pi-pi/3`

`{4pi}/3 < 2x < {5pi}/3`

`{2pi}/3 < x < {5pi}/6`

After that we combine the two solution sets together (union) and we get for `x in [0,pi)`, the solution set is

`(pi/6,pi/3) uu ({2pi}/3,{5pi}/6)`.

For `x in RR`, you just have to add integer multiples of `pi`. So the final answer looks like

`... uu (-{5pi}/6,-{2pi}/3) uu (-pi/3,-pi/6) uu(pi/6,pi/3) uu ({2pi}/3,{5pi}/6) uu ({7pi}/6,{4pi}/3) uu ({5pi}/3,{11pi}/6) uu ...`

Answer 2

(pi/6, pi/3)
(2pi/3, (5pi)/6)

Explanation:

Separate the inequality in 2 parts:
a. `sin 2x > sqrt3/2` and
b. `- sin 2x > sqrt3/2` --> `sin 2x < -sqrt3/2`
a. Solve `sin 2x > sqrt3/2` by the trig unit circle.
The inequality is true when the arc 2x varies between `pi/3` and `(2pi)/3` on the unit circle
`pi/3 < 2x < (2pi)/3`
`pi/6 < x < (2pi)/6 = pi/3`
b. Solve `sin 2x < - sqrt3/2`.
The inequality is true when the arc 2x varies between `(4pi)/3` and `(5pi)/3:` on the unit circle. In this interval, `sin 2x < - sqrt3/2`
`(4pi)/3 < 2x < (5pi)/3`
`(2pi)/3 < x < (5pi)/6`