
Mark M. answered 12/08/21
Mathematics Teacher - NCLB Highly Qualified
2 sin θ - 1 = csc θ
2 sin θ - 1 = 1 / sin θ
2 sin2 θ - sin θ = 1
Use Algebraic techniques to solve for θ.
Lin L.
asked 12/08/21Solve each equation for exact solutions over the interval [00, 3600].
2sinθ−1=cscθ
Select one:
a.
{300, 2100, 2400, 3000}
b.
{900, 2100, 3300}
c.
{00, 450, 2250}
d.
{300, 2000, 3100}
e.
{150, 1300, 4300}
Mark M. answered 12/08/21
Mathematics Teacher - NCLB Highly Qualified
2 sin θ - 1 = csc θ
2 sin θ - 1 = 1 / sin θ
2 sin2 θ - sin θ = 1
Use Algebraic techniques to solve for θ.
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