What is the general solution for arctan(e^x)= pi/4 ?

The general solution for arctan(e^x)= pi/4 is x=0

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arctan(e^x)= pi/4

arctan (e^{x}) = \frac{π}{4}

x = 0

Solution steps

arctan (e^{x}) = \frac{π}{4}

Apply trig inverse properties

arctan (e^{x}) = \frac{π}{4}

arctan (x) = a \Rightarrow x = tan (a)

e^{x} = tan (\frac{π}{4})

tan (\frac{π}{4}) = 1

tan (\frac{π}{4})

Use the following trivial identity:

tan (\frac{π}{4}) = 1

tan (\frac{π}{4})

tan (x)

periodicity table with

πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

= 1

= 1

e^{x} = 1

e^{x} = 1

Solve

e^{x} = 1 : x = 0

e^{x} = 1

Apply exponent rules

e^{x} = 1

f (x) = g (x)

, then

ln (f (x)) = ln (g (x))

ln (e^{x}) = ln (1)

Apply log rule:

ln (e^{a}) = a

ln (e^{x}) = x

x = ln (1)

Simplify

ln (1) : 0

ln (1)

Apply log rule:

lo g_{a} (1) = 0

= 0

x = 0

x = 0

x = 0

csc^{2} (x) + 2 cos (x) + 1 = 0

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

tan (t) = - \frac{12}{5}, \frac{3 π}{2} < t < 2 π

3 sin (x) + 5 cos (x) = 4

- 2 = sec (x)

What is the general solution for arctan(e^x)= pi/4 ?
The general solution for arctan(e^x)= pi/4 is x=0