determine the values of theta where sin theta =-3square root over 2 and 0 greater/equal sign thethat greater/equal 2pi
the question should be read as
where does sinθ = -√3/2 where 0 <= θ < 2π
we know that sin π/3 = √3/2, and
sinθ < 0 in QIII and QIV
so sinθ = -√3/2 if θ = 4π/3 or 5π/3
The square root of 2 is a positive number, so it is not possible for the sine of an angle to equal -3 times the square root of 2. Therefore, there are no values of theta that satisfy the given condition.
Apologies for the misunderstanding.
Indeed, we can consider the unit circle to find the values of theta where sin theta = -√3/2, where 0<=theta<2π.
Recall that sin(pi/3) = √3/2.
On the unit circle, sin theta corresponds to the y-coordinate of a point on the circle. Since sin theta is negative (sin theta < 0) in Quadrant III and Quadrant IV, we need to find the angles in these quadrants that have the same absolute value as sin(pi/3).
In Quadrant III, the angle theta that satisfies sin theta = -√3/2 is θ = 2π - π/3 = 5π/3.
In Quadrant IV, the angle theta that satisfies sin theta = -√3/2 is θ = 2π + π/3 = 7π/3.
Therefore, the values of theta where sin theta = -√3/2, and 0<=theta<2π, are θ = 5π/3 and θ = 7π/3.