h_finding solutions in an interval for a trigonometric equation with an angle_a
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8/20/2019 H_Finding Solutions in an Interval for a Trigonometric Equation With an Angle_a
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Kwadwo Amankwa - 03/01/2016 3:01:52 AM PSTPreCalculus / Amankwa, Kwadwo (Amankwa)
Finding solutions in an interval for a trigonometric equation with an anglemultiplied by a constant
Find all solutions of the equation in the interval .
Write your answer in radians in terms of .If there is more than one solution, separate them with commas.
Because is being multiplied by a constant other than , we must be careful when
finding solutions in an interval.
We'll first find the general formulas giving all solutions. Then we'll compute the
particular solutions in .
Since sine has a period of , we want to find the points in the interval where
sine has the value .
The only such points are and , which give the equations and .
To get the general formulas giving all the solutions of , we add multiples of
.
or
Substituting different values for in the above formulas, we can find the solutions in
.
Note that any negative gives a solution less than , so we start with .
0, 2π
=sin2 x
3 0
π
x 1
0, 2π
2π 0, 2π
0
0 π =2 x
3 0 =
2 x
3 π
=sin2 x
3 0
2π
2 x
3= +0 2k π
2 x
3= +π 2k π , k
x = 3k π x = +3π
2 3k π
k
0, 2π
k 0 =k 0
=k :0 = x =3 0 π 0
= x =+3π
2 3 0 π
3π
2
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Not in
Not in
The solutions in are and .
The answer is , .
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=k :1 = x =3 1 π 3π 0, 2π
= x =+3π
2 3 1 π =+
3π
2
6π
2
9π
20, 2π
0, 2π = x 0 = x
3π
2
= x 03π
2