Question

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

Answer 1

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

Explanation:

`cos 2x = 1 - 2sin^2 x`. The equation becomes:
`2sin^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: sin x = - 1 and `sin x = -c/a = 1/2`.
a. sin x = -1 --> `x = (3pi)/2`
b. `sin x = 1/2`.
Trig unit circle --> There are 2 arcs x that have same sin value `(1/2)` -->
`x = pi/6 and x = pi - pi/6 = (5pi)/6`
For interval `(0, 2pi)`, there are 3 ansawers:
`pi/6, (5pi)/6, (3pi)/2`