Question

How do you solve `sin^2 x- 8 sin x - 4= 0` and find all values of x in the interval`[0, 360^o)`?

Answer 1

`f(x) = sin^2 x - 8sin x - 4 = 0`. Call `sin x = t`, we have:
`f(t) = t^2 - 8t - 4`. This is a quadratic equation. Solve it by using the new quadratic formula #d^2 = b^2 - 4ac = 64 + 16 = 80 = 16*5 rArr d = 4*sqrt(5)#.
`x_1 = -b/(2a) + d/(2a) = 8/2 + (4*sqrt(5))/2 = 4 + 2sqrt(5) approx 8.472` (rejected)
`x_2 = -b/(2a) - d/(2a) = 4 - 2sqrt(5) approx -0.47`
Next, solve `t = sin x = -0.472`. Calculator give `x_1 = 180 + 28.16 = 208.16°`.
Trig unit circle gives another arc `x` that has the same sine value (`-0.472`).
`x_2 = 360 - 28.16 = 331.84°`
Answers within interval `(0°, 360°)`: `x_1 = 208.16°`; `x_2 = 331.84°`.
Check
`x = 208.16° rArr sin x = -0.472 rArr f(x) = (-0.472)^2 - 8(-0.472) - 4 = 0.22 + 3.78 - 4 = 0` Correct.

Answer 2

I propose a slightly (but not too) different way of obtaining the solution (approximation in the very last step).

We start from `sin^2 x−8 sin x−4=0` and the first step, as in Nghi's answer, is to substitute `t=sin x`. We get `t^2-8t-4=0`, which is a quadratic formula whose solutions are

`t_1=frac{-(-8)+sqrt{(-8)^2-4*1*(-4)}}{2*1}=4+2sqrt{5}`
`t_2=frac{-(-8)-sqrt{(-8)^2-4*1*(-4)}}{2*1}=4-2sqrt{5}`

To get the solution of the initial equation, we have to find all `x_1 in [0°,360°)` such that `sin x_1=4+2sqrt{5}` and all `x_2 in [0°,360°)` such that `sin x_2 =4-2sqrt{5}`.

It turns out that `sin x_1=4+2sqrt{5}` has no solution `x_1`, since `sin x_1 in [-1,1]` for all `x_1 in [0°,360°)` and `4+2sqrt{5}>1`.

On the other hand, `sin x_2 =4-2sqrt{5}` has at least a solution: it's easy to check that `-1<4-2sqrt{5}<0`.
graph{sin(x) [-0.1, 6.4, -1.1, 1.1]}
The graph of the sine function suggests that the solutions will be two, both in the interval `[180°,360°)`.

So we get
`x_{2,1}=180°-arcsin(4-2sqrt{5})`
`x_{2,2}=360°+arcsin(4-2sqrt{5})`

Since `arcsin(4-2sqrt{5}) approx -28.173°` we get that `x_{2,1} approx 208.173 °`and `x_{2,2} approx 331.827 °`.