What is the general solution for arccos(1-x)+arccos(x)=arccos(-x) ?

The general solution for arccos(1-x)+arccos(x)=arccos(-x) is x=0,x= 1/2

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arccos(1-x)+arccos(x)=arccos(-x)

arccos (1 - x) + arccos (x) = arccos (- x)

x = 0, x = \frac{1}{2}

Solution steps

arccos (1 - x) + arccos (x) = arccos (- x)

a = b \Rightarrow cos (a) = cos (b)

cos (arccos (1 - x) + arccos (x)) = cos (arccos (- x))

Use the following identity:

cos (s + t) = cos (s) cos (t) - sin (s) sin (t)

cos (arccos (1 - x)) cos (arccos (x)) - sin (arccos (1 - x)) sin (arccos (x)) = cos (arccos (- x))

Use the following identity:

cos (arccos (x)) = x

Use the following identity:

cos (arccos (x)) = x

Use the following identity:

sin (arccos (x)) = 1 - x^{2}

Use the following identity:

sin (arccos (x)) = 1 - x^{2}

(1 - x) x - 1 - (1 - x)^{2} 1 - x^{2} = - x

Solve

(1 - x) x - 1 - (1 - x)^{2} 1 - x^{2} = - x : x = 0, x = \frac{1}{2}

x = 0, x = \frac{1}{2}

Verify solutions by plugging them into the original equation

x = 0, x = \frac{1}{2}

tan (x) = 0.75

2 sin (x) tan (x) = tan (x)

4 csc (θ) + 5 = 0

cos (x) = - cos (x) + 2

4 cos^{2} (2 x) - 3 = 0

What is the general solution for arccos(1-x)+arccos(x)=arccos(-x) ?
The general solution for arccos(1-x)+arccos(x)=arccos(-x) is x=0,x= 1/2