Can anyone help explain/show how to solve this? Half angle trig identity

Question

Solve the following equation using half angle identities, given that x is in the interval [0, 360]. Choose all values that apply. Your answers should be expressed in degrees, to the nearest tenths.

8tan(x/2) + 8cosxtan(x/2) = 1

Answer 1

tan(x/2) = (1 - cosx) / sinx

So, 8(1 - cosx)/sinx + 8cosx(1 - cosx)/sinx - 1 = 0

8/sinx - 8cosx/sinx + 8cosx/sinx - 8cos²x/sinx - 1 = 0

8/sinx -8cos²x/sinx - sinx / sinx = 0

-8cos²x - sinx + 8 = 0

-8(1 - sin²x) - sinx + 8 = 0

8sin²x - sinx = 0

sinx (8sinx - 1) = 0

sinx = 0 or sinx = 1/8

If sinx = 0 rhen x = 0°, 180°, or 360° (these values make the original equation invalid)

So, sinx = 1/8

x = Sin^-1(1/8) or 180° - Sin^-1(1/8)

Answer 2

Answer 3

I would start by factoring out the 8tan(x/2):

8[tan(x/2)][1+cos(x)] = 1

Now replace the tan(x/2) with its equivalent:

8 √[(1-cos(x))/(1+cos(x))](1+cos(x)) = 1 (ignoring the ±on the half angle identity for now)

Since this equation has a radical, isolate the radical then square both sides:

√[(1-cos(x))/(1+cos(x))] = 1/[8(1+cos(x)]

(1-cos(x))/(1+cos(x)) = 1/[64(1+cos(x)²] :(this step might have introduced an extraneous root)

(1+cos(x))²(1-cos(x))/(1+cos(x)) = 1/64

(1+cos(x))(1-cos(x)) = 1/64

1-cos²(x) = 1/64

cos²(x) = 63/64

cos(x) = ±3√7 / 8

x₁ = arccos(3√7 /8)

x₂=arccos(-3√7/8)

Since the question asks for answer in degrees, make sure your calculator is not in radians.

Need to confirm those answers were not introduced as extraneous roots.

Use the following Desmos graph to confirm.

desmos.com/calculator/kc71c9bvt0

When you evaluate arccos, you can also find some angles in the 3rd and 4th quadrants, but those values do not check when you evaluate in the original equation.