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

#|sin2x|>sqrt3/2 #

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#