SOLUTION: help! Question Find the exact solution to the equation in the interval [3π/2, 5π/2]. sin (t) = (-squareroot3/2)

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Question 1003534: help!


Question Find the exact solution to the equation in the interval
[3π/2, 5π/2].
sin (t) = (-squareroot3/2)




Found 2 solutions by ikleyn, stanbon:
Answer by ikleyn(52812) About Me  (Show Source):
You can put this solution on YOUR website!
.
Find the exact solution to the equation in the interval [3π/2, 5π/2]:
sin (t) = (-squareroot3/2).
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sin(t) = -sqrt%283%29%2F2.

The solutions in the interval  [0, 2pi)  are

t = 4pi%2F3  and  %285pi%29%2F3.

Of these two only one is in the interval  [3pi%2F2, 5pi%2F2].

It is  t = %285pi%29%2F3.



Plot  y = sin(x)  and  y = -sqrt%283%29%2F2
in the interval  [3pi%2F2, 5pi%2F2].


Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
Question Find the exact solution to the equation in the interval
[3π/2, 5π/2].
sin (t) = (-sqrt(3)/2))
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sin(t) has that value in QIII and QIV
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The reference angle is pi/3
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In QIII t = (4/3)pi
In QIV t = (5/3)pi
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In the interval [(3/2)pi,(5/2)pi] t = (5/3)pi
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Cheers,
Stan H.
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