What is the general solution for -4arctan(x)=pi ?

The general solution for -4arctan(x)=pi is x=-1

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-4arctan(x)=pi

- 4 arctan (x) = π

x = - 1

Solution steps

- 4 arctan (x) = π

Divide both sides by

- 4

- 4 arctan (x) = π

Divide both sides by

- 4

\frac{- 4 arctan ( x )}{- 4} = \frac{π}{- 4}

Simplify

arctan (x) = - \frac{π}{4}

arctan (x) = - \frac{π}{4}

Apply trig inverse properties

arctan (x) = - \frac{π}{4}

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

x = tan (- \frac{π}{4})

tan (- \frac{π}{4}) = - 1

tan (- \frac{π}{4})

Use the following property:

tan (- x) = - tan (x)

tan (- \frac{π}{4}) = - tan (\frac{π}{4})

= - 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

x = - 1

x = - 1

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

sin (x) = - \frac{3}{2}, 0 \leq x \leq 2 π

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

cos^{2} (θ) = 0

cos (7 x) cos (2 x) + sin (7 x) sin (2 x) = \frac{1}{2}

What is the general solution for -4arctan(x)=pi ?
The general solution for -4arctan(x)=pi is x=-1