Use inverse functions where needed to find all solutions of the equation in the interval [0,2pi). 8sin^2x+6cosx-9=0

Question

This involves solving Trigonometric equations

Answer 1

Since this is a quadratic equation, might be able to solve by factoring. But first we need to write with only one trig function. Since cosx is the linear term, consider replacing sin²(x) with (1-cos²x), and then converting to standard form.

8(1-cos²x) + 6cosx - 9 = 0

8 -8cos²x + 6cosx - 9 = 0

8cos²x -6cosx +1 = 0 (after combining similar terms, multiply through by -1 so leading coefficient is positive--making it easier to factor)

(4 cosx - 1) (2cosx -1 ) = 0

x = cos^-1(1/4) or x = cos^-1(1/2)

Those will each give two solutions in the interval [0, 2pi). Two of the solutions will be between 0 and pi/2. The other two will be between 3pi/2 and 2pi.