-5sin^2(t)-4cos(t)=-4

Question

-5sin^2(t)-4cos(t)=-4

Find all solutions to equation below in the interval [0,2pi)

Answer 1

-5 sin² (t) - 4 cos (t) = 4

-5(1 - cos² (t)) - 4 cos (t) - 4 = 0

-5 + 5 cos²(t) - 4 cos (t) - 4 = 0

5 cos²(t) - 4 cos (t) - 9 = 0

(5 cos (t) - 9)(cos (t) + 1) = 0

Can you solve for t and answer?

Answer 2

Substitute 1-cos²t for sin²t and the equation becomes a quadratic is cos t.

Solve the quadratic for cos t and then the solution will be arccos.

But be careful, there may be extraneous roots and/or multiple values of x which will satisfy.

Obviously x=0 and x=2pi are roots and there are 2 more roots in the interval.