A class was asked to solve 3 cos = 2 sin for 0 <

A class was asked to solve 3 cos θ = 2 − sin θ for 0 < θ, 360°. One student expressed the equation in the form R cos (θ − α) = 2, with R > 0 and 0 < α < 90°, and correctly solved the equation.

a. Find the values of R and α and hence find her solutions. Another student decided to square both sides of the equation and then form a quadratic equation in sin θ.

b. Show that the correct quadratic equation is 10 sin² θ − 4 sin θ − 5 = 0.

c. Solve this equation, for 0 < θ, 360°.

d. Explain why not all of the answers satisfy 3 cos θ = 2 − sin θ.