Question

Question #d89db

Answer 1

`pi/6; (5pi)/6`

Explanation:

`f(x) = 4cos^2 x - 4sin x - 1 = 0`
Replace `cos^2 x` by `(1 - sin^2 x)`
`4(1 - sin^2 x) - 4sin x - 1 = 0`
`4 - 4sin^2 x - 4sin x - 1 = 0`
Change side. Solve the quadratic equation by the improved quadratic formula in graphic form (Socratic, Google Search):
`4sin^2 x + 4sin x - 3 = 0`
`D = d^2 = b^2 - 4ac = 16 + 48 = 64` --> `d = +- 8`

There are 2 real roots:
`sin x = - b/(2a) +- d/(2a) = - 1/2 +- 1`
`sin x = - 1/2 + 1 = 1/2` and `sin x = - 3/2` (rejected as < - 1)
`sin x = 1/2` --> trig table and unit circle give -->
`x = pi/6`, and `x = pi - pi/6 = (5pi)/6`