How can i solve this (trigonometry)?

#(sinx-1/2)(sinx-2)<=0# over the interval #[0;2pi]#

#(sinx-1/2)(sinx-2)<=0# over the interval #[0;2pi]#

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Nov 17, 2016

Answer:

#x in [pi/6, (5pi)/6]#

Explanation:

Alternate answer, using the quadratic formula, as requested.

First, find where the left hand side is equal to #0# using the quadratic formula:

#(sinx-1/2)(sinx-2) = 0#

#=> (sinx)^2-5/2sinx + 1 = 0#

#=> sinx = (-(-5/2)+-sqrt((-5/2)^2-4(1)(1)))/(2(1))#

#=(5/2+-sqrt(9/4))/2#

#=(5/2+-3/2)/2#

#=5/4+-3/4#

#=> sinx = 2 or sinx = 1/2#

(Note that we could have arrived at this conclusion directly from the factored form by setting each factor equal to #0#)

As #sinx != 2# for any #x in RR#, this leaves us with

#sinx = 1/2#

Using knowledge of common angles, a unit circle, or a calculator, we find that #sinx = 1/2# has the solutions #x in {pi/6, (5pi)/6}# in the interval #[0, 2pi]#.

As #(sinx-1/2)(sinx-2)# is a continuous function, any interval in which it changes sign must include a #0#. Thus, if we split the interval #[0, 2pi]# by removing the points #{pi/6, (5pi)/6}#, we get three intervals:

#[0, pi/6)#, #(pi/6, (5pi)/6)#, #((5pi)/6, 2pi]#

and the sign of #(sinx-1/2)(sinx-2)# is constant on each of these intervals, as none of them contain a #0# of #(sinx-1/2)(sinx-2)#.

With that, we can determine the sign of #(sinx-1/2)(sinx-2)# on each interval by testing a value in each one, and noting that the sign of that value will be the same as the sign of all of the values on that interval.


#0 in [0, pi/6)# and

#(sin(0)-1/2)(sin(0)-2) = (-1/2)(-2) = 1 > 0#

Thus #(sinx-1/2)(sinx-2) > 0# on #[0, pi/6)#


#1 in (pi/6, (5pi)/6)# and

#(sin(1)-1/2)(sin(1)-2) = (1/2)(-1) = -1/2 < 0#

Thus #(sinx-1/2)(sinx-2) < 0# on #(pi/6, (5pi)/6)#


#pi in ((5pi)/6, 2pi]# and

#(sin(2pi)-1/2)(sin(2pi)-2) = (-1/2)(-2) = 1 > 0#

Thus #(sinx-1/2)(sinx-2) > 0# on #((5pi)/6, 2pi]#


Adding in the points at which the function is #0#, we obtain our final result:

#(sinx-1/2)(sinx-2) <= 0# if #x in [pi/6, (5pi)/6]#

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Nov 16, 2016

Answer:

#x in [pi/6, (5pi)/6]#

Explanation:

Note first that as #-1 <= sinx <= 1# for all real #x#, we must have #sinx-2 <= -1 < 0#. Then, #(sinx-1/2)(sinx-2)# is the product of #sinx-1/2# and a negative number, meaning it is less than or equal to #0# if and only if #sinx-1/2 >= 0#.

Thus, the given problem is equivalent to the problem #sinx-1/2 >= 0# with the restriction #x in [0, 2pi]#

Adding #1/2# to each side of the inequality, we get

#sinx >= 1/2#

Using the unit circle, we can tell that on our restricted interval, we have #sinx>=1/2# if and only if #x in [pi/6, (5pi)/6]#. Therefore, this is our answer.

#x in [pi/6, (5pi)/6]#

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