Question

2cos^2x-sinx=-1?

Answer 1

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

Explanation:

Solve the equation:
`2cos^2 x - sin x + 1 = 0`
Replace in the equation `cos^2 x` by `(1 - sin^2 x)` -->
`2 - 2sin^2 x - sin x - 1 = 0`
Solve this quadratic equation for sin x -->
`-2sin^2 x - sin x + 1 = 0`
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
Unit circle gives --> `x = (3pi)/2 + 2kpi`
b. `sin x = 1/2`
Trig table and unit circle give 2 solutions for x -->
`x = pi/6`, and `x = (5pi)/6`