What is the general solution for arccos(x)+arccos(2x)= pi/2 ?

The general solution for arccos(x)+arccos(2x)= pi/2 is x= 1/(sqrt(5))

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arccos(x)+arccos(2x)= pi/2

arccos (x) + arccos (2 x) = \frac{π}{2}

x = \frac{1}{5}

Solution steps

arccos (x) + arccos (2 x) = \frac{π}{2}

Rewrite using trig identities

arccos (x \cdot 2 x - (1 - x^{2}) (1 - (2 x)^{2})) = \frac{π}{2}

Apply trig inverse properties

x \cdot 2 x - (1 - x^{2}) (1 - (2 x)^{2}) = 0

Solve

x \cdot 2 x - (1 - x^{2}) (1 - (2 x)^{2}) = 0 : x = \frac{1}{5}, x = - \frac{1}{5}

x = \frac{1}{5}, x = - \frac{1}{5}

Verify solutions by plugging them into the original equation

x = \frac{1}{5}

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

- 2 - 3 sin (x) = 1 - 6 sin (x)

x, 4 cos (x) sin (y) = 1

- 2 csc^{2} (x) = 0

cos (x) = - 0.2

What is the general solution for arccos(x)+arccos(2x)= pi/2 ?
The general solution for arccos(x)+arccos(2x)= pi/2 is x= 1/(sqrt(5))