What is the general solution for arctan(x/3)+arctan(x/2)=arctan(x) ?

The general solution for arctan(x/3)+arctan(x/2)=arctan(x) is x=0,x=-1,x=1

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arctan(x/3)+arctan(x/2)=arctan(x)

arctan (\frac{x}{3}) + arctan (\frac{x}{2}) = arctan (x)

x = 0, x = - 1, x = 1

Solution steps

arctan (\frac{x}{3}) + arctan (\frac{x}{2}) = arctan (x)

Subtract

arctan (x)

from both sides

arctan (\frac{x}{3}) + arctan (\frac{x}{2}) - arctan (x) = 0

Rewrite using trig identities

arctan \frac{\frac{\frac{x}{3} + \frac{x}{2}}{1 - \frac{x}{3} \cdot \frac{x}{2}} - x}{1 + \frac{\frac{x}{3} + \frac{x}{2}}{1 - \frac{x}{3} \cdot \frac{x}{2}} x} = 0

Apply trig inverse properties

\frac{\frac{\frac{x}{3} + \frac{x}{2}}{1 - \frac{x}{3} \cdot \frac{x}{2}} - x}{1 + \frac{\frac{x}{3} + \frac{x}{2}}{1 - \frac{x}{3} \cdot \frac{x}{2}} x} = 0

Solve

\frac{\frac{\frac{x}{3} + \frac{x}{2}}{1 - \frac{x}{3} \cdot \frac{x}{2}} - x}{1 + \frac{\frac{x}{3} + \frac{x}{2}}{1 - \frac{x}{3} \cdot \frac{x}{2}} x} = 0 : x = 0, x = - 1, x = 1

x = 0, x = - 1, x = 1

Verify solutions by plugging them into the original equation

x = 0, x = - 1, x = 1

w, y = arctan (1 + 4 w)

cos (8 x) = 1

cos^{4} (a) = 8 cos^{4} (a) - 8 cos^{2} (a) + 1

sin^{2} (a) - 4 sin (a) + 3 = 0

4 sin^{2} (x) - 4 cos (x) - 1 = 0

What is the general solution for arctan(x/3)+arctan(x/2)=arctan(x) ?
The general solution for arctan(x/3)+arctan(x/2)=arctan(x) is x=0,x=-1,x=1