Solve for 0<x≤2π, cos^2(x)-3cos(x)=5sin^2(x)-2, giving all answers exactly

Use the identity cos^2(x)+sin^2(x)=1, to change the 5sin^(x) into 5-5cos^2(x) and rearrange. Let y=sin(x) and solve the rearranged quadratic (6y^2-3y-3) for y and find the value of x for each solution. y= -0.5, 1 leads to x=2π/3, 4π/3, 2π, explained with cast diagram if needs be.

Answered by Ben M. Maths tutor

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