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

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

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arccos(x)-arcsin(x)=arcsin(1-x)

arccos (x) - arcsin (x) = arcsin (1 - x)

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

Solution steps

arccos (x) - arcsin (x) = arcsin (1 - x)

a = b \Rightarrow sin (a) = sin (b)

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

Use the following identity:

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

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

Use the following identity:

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

Use the following identity:

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

Use the following identity:

cos (arccos (x)) = x

Use the following identity:

sin (arcsin (x)) = x

1 - x^{2} 1 - x^{2} - xx = 1 - x

Solve

1 - x^{2} 1 - x^{2} - xx = 1 - 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}

sin^{2} (x) = ∣ sin (x) ∣

1 - cos^{2} (x) - sin^{22} (x) = 0

cos (\frac{x}{4}) sin (\frac{x}{4}) = 3 sin (\frac{x}{4}) cos (\frac{x}{4})

5 cos (x) = 1 + 2 sin^{2} (x)

tan (2 x + 1) = - cot (x + 3)

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