How to solve this inequality |sin2x|>sqrt3/2 |sin2x|>32 ?

|sin2x|>sqrt3/2 |sin2x|>32

2 Answers
Jan 31, 2016

... 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 ...

Jan 31, 2016

(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