
Mark M. answered 04/17/19
Mathematics Teacher - NCLB Highly Qualified
cos 2θ = 1 - sin2 θ
6 sin2 θ - 5 cos 2θ = 0
6 sin2 θ - 5(1 - sin2 θ) = 0
6 sin2 θ - 5 + 5 sin2 θ = 0
11 sin2 θ = 5
Can you continue and solve for θ?
Shay L.
asked 04/17/19Find all degree solutions in the interval 0° ≤ θ < 360°. If rounding is necessary, round to the nearest tenth of a degree. Use your graphing calculator to verify the solution graphically. (Enter your answers as a comma-separated list.)
6 sin2 θ − 5 cos 2θ = 0
Mark M. answered 04/17/19
Mathematics Teacher - NCLB Highly Qualified
cos 2θ = 1 - sin2 θ
6 sin2 θ - 5 cos 2θ = 0
6 sin2 θ - 5(1 - sin2 θ) = 0
6 sin2 θ - 5 + 5 sin2 θ = 0
11 sin2 θ = 5
Can you continue and solve for θ?
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