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

2 Answers

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.

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 °.