Question

How do you solve `cos 2 x = sin x` over the interval 0 to 2pi?

Answer 1

`pi/6, (5pi)/6 and (3pi)/2`

Explanation:

Apply the trig identity: `cos 2x = 1 - 2sin^2 x.` We get:
`1 - 2sin^2 x = sin x`
`- 2cos^2 x - sin x + 1 = 0.`
Solve this quadratic equation for sin x.
Since a - b + c = 0, use shortcut. The 2 real roots are; -1 and
`-c/a = 1/2.`
Trig unit circle and trig table give -->
a. sin x = -1 -->`x = (3pi)/2.`
b. `sin x = 1/2` --> `x = pi/6` and `x = pi - pi/6 = (5pi)/6`