Solve arccos x+arccos 2x=60 degrees?

Question

what I did is arccosx=a cos a= x sin a=sqroot1+xsquared
and for arccos 2x i let it equal b so cosb=2x < is that correct i'm not sure about that. and sin b=sqroot 1+4xsquared and the next step would be using a compound formula but would i use cos(a+b) or something else. because in an example there was arcsin x+ arcsin 2x= 60 degrees and the compound fomrula they used was cos(a+b)

Answer 1

Solve arccos(x)+arccos(2x) = 60°

Let arccos(x) = θ
So cosθ = x
And sinθ = ±√(1 - x²)
(as sin²θ + cos²θ = 1 ie sinθ = cos²θ)

and arcos(2x) = φ
So cosφ = 2x
And sinφ = ±√(1 - 4x²)
(as sin²φ + cos²φ = 1 ie sinφ = ±√(1 - cos²φ)

Let therefore θ + φ = 60°

So cos(θ + φ) = cos60° = ½

so cosθcosφ + sinθsinφ = ½

ie x*2x ± √(1 - x²)*√(1 - 4x²) = ½

ie 2x² ± √(1 - 5x² +4x^4) = ½

So ± 2√(1 - 5x² +4x^4) = 1 - 4x²

Square both sides
4(1 - 5x² +4x^4) = 1 - 8x² + 16x^4

So 4 - 20x² + 16x^4 = 1 - 8x² + 16x^4

Thus 12x² = 3
x² = ¼
x = ± ½

Check:

x = ½ and so 2x = 1
arccos(½) = 60°
arccos(1) = 0°
60° + 0° = 60° Yess!!!

x = -½
arccos -½ = 120°
arccos -1 = 180°
120° + 180° = 300° Noooo!!!!

So the only solution is x = ½ (x = -½ is an extraneous solution arising from the squaring step and also the squaring and the squre root steps way above ie sina = ±√(1 -cos²a))

However you can get the solution x = -½ to work if you are allowed to have a broader range of outputs for arcos x other than 0° ≤ x ≤ 180°

After all if cos A = -½ then A can equal 240° and if cos B = -1 then A can equal -180° and then 240° + -180° = 60°

Answer 2

Your notation is a little hard to follow, let me work on this a bit and be back with you...
let a = arccos x and b = arccos 2x so that cos a = x and cos b = 2x
So now you have a+ b = 60 degrees and cos (a + b) = cos (60 degrees).
Using the addition formula on the right, we get
cos a cos b - sin a sinb which is
2x^2 - sqrt ((1-x^2)(1-4x^2)) = cos 60 = 1/2
To get the second part, remember that sin a = sqrt (1-cos^2a) and sin b = sqrt(1 - (2x)^2)
Simplify the left side, solve for x, then extend this to the unit circle for all solutions between 0 and 360, or all possible solutions.

Answer 3

Finally you get x=1/2

Answer 4

i could go away it as arccot (a million/x) via fact the identity makes use of cot. cot (a + b) = (cot b cot a - a million)/(cot b + cot a) If an perspective has a cos of x, its cot is x / ?(a million - x^2). So plug those in. (a million/x • x/?(a million - x^2)) - a million over (a million/x + x/?(a million - x^2) then simplify if mandatory perhaps? i'm no longer a hundred% helpful... see what you think of.