Question

How do you solve `4sin^2x + 1 = -4sin x`?

Answer 1

`x= pi/6 +k*2pi`
or
`x= (5pi)/6 + k*2pi`
`color(white)("XXXX")` `color(white)("XXXX")`for all `k epsilon ZZ`

Explanation:

`4sin^2(x) + 1 = 4 sin(x)`

`rArr 4sin^2(x)-4sin(x)+1=0`

`rArr (2sin(x)-1)^2 = 0`

`rArr 2sin(x) = 1`

`rArr sin(x) = 1/2`

If we limit the range of `x` to `[0,2pi]`
the above represents a standard trigonometric triangle with
`color(white)("XXXX")` `x = pi/6 or (5pi)/6`