We can translate the equation sinv=−12 to the problem of finding those angles in the unit circle which have a y-coordinate of −12. If we compare this with the problem that we had in exercise a, where we looked for angles which satisfied sinv=+12, then the situation is the same, except that the angles now lie under, rather than above, the x-axis, due to reflectional symmetry.
Angle 2π - π/6 = 11π/6
Angle π + π/6 = 7π/6
The two angles which satisfy sinv=−12 lie in the third and fourth quadrants and are v=2−6=116 and v=+6=76.
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