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

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#|sin2x|>sqrt3/2 #

##### 2 Answers

#### Explanation:

We can say more generally that we want to find the values of

#sqrt3/2 < sin2x# or#sin2x < sqrt3/2# .

The period of

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

Therefore, we will limit

We first attempt to solve the first inequality.

#sqrt3/2 < sin2x#

Since

For the second inequality,

#sin2x < sqrt3/2#

Since

After that we combine the two solution sets together (union) and we get for

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

For

#... 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 ...#

(pi/6, pi/3)

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

#### Explanation:

Separate the inequality in 2 parts:

a.

b.

a. Solve

The inequality is true when the arc 2x varies between

b. Solve

The inequality is true when the arc 2x varies between