Question

How do you find all real numbers on the interval [0,2pi) that satisfy the equation and use radian measure `2sin^2(x)-3sin(x)=-1`?

Answer 1

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

Explanation:

`f(x) = 2sin^2 x - 3sin 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 = pi/2`
b. `sin x = 1/2.` The trig unit circle gives 2 arcs that have the same sin value (1/2)
`x = pi/6 and x = pi - pi/6 = (5pi)/6`