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(52814) (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) = .
The solutions in the interval [0, ) are
t = and .
Of these two only one is in the interval [, ].
It is t = .
Plot y = sin(x) and y = -
in the interval [, ].
Answer by stanbon(75887) (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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