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Given 0=<x<2pi find the roots of the equation sinx=-1/2.

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We have to find the values of x for which sin x = -1/2 in the range [0, 2*pi)

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giorgiana1976 | Student

In other words, we'll have to determine all possible values for x, located in the range [0,2pi), that makes the value of sine function -1/2.

The sine function is negative in the 3rd and the 4th quadrants.

x = arcsin(-1/2)

x = pi + pi/6

x = 7pi/6 (3rd quadrant)

x = 2pi - pi/6

x = 11pi/6 (4th quadrant)

The only possible values for x, located in the range [0,2pi), are: {7pi/6 ; 11pi/6}.

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