What is the general solution for 2^{arccos(x)}+4^{arccos(x)+1}-2=3 ?

The general solution for 2^{arccos(x)}+4^{arccos(x)+1}-2=3 is x=1

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2^{arccos(x)}+4^{arccos(x)+1}-2=3

2^{a r c c o s (x)} + 4^{a r c c o s (x) + 1} - 2 = 3

x = 1

Solution steps

2^{a r c c o s (x)} + 4^{a r c c o s (x) + 1} - 2 = 3

Solve by substitution

arccos (x) = 0

arccos (x) = 0 : x = 1

Combine all the solutions

x = 1

csc (θ) cos (θ) sin (θ) = 1

3 cos (x) + sin (2 x) = 0

tan (α) = 0.875

sin (x) = \frac{18}{31}

4 cos^{3} (x) - 3 cos (x) = - 0.7071

What is the general solution for 2^{arccos(x)}+4^{arccos(x)+1}-2=3 ?
The general solution for 2^{arccos(x)}+4^{arccos(x)+1}-2=3 is x=1