What is the general solution for solvefor x,tan^2(x)=1 ?

The general solution for solvefor x,tan^2(x)=1 is x= pi/4+pin,x=(3pi)/4+pin

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solvefor x,tan^2(x)=1

solve for

x, tan^{2} (x) = 1

x = \frac{π}{4} + πn, x = \frac{3 π}{4} + πn

Degrees

x = 4 5^{\circ} + 18 0^{\circ} n, x = 13 5^{\circ} + 18 0^{\circ} n

Solution steps

tan^{2} (x) = 1

Solve by substitution

tan^{2} (x) = 1

Let:

tan (x) = u

u^{2} = 1

u^{2} = 1 : u = 1, u = - 1

u^{2} = 1

For

x^{2} = f (a)

the solutions are

x = f (a), - f (a)

u = 1, u = - 1

1 = 1

1

Apply rule

1 = 1

= 1

- 1 = - 1

- 1

Apply rule

1 = 1

= - 1

u = 1, u = - 1

Substitute back

u = tan (x)

tan (x) = 1, tan (x) = - 1

tan (x) = 1, tan (x) = - 1

tan (x) = 1 : x = \frac{π}{4} + πn

tan (x) = 1

General solutions for

tan (x) = 1

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}

x = \frac{π}{4} + πn

x = \frac{π}{4} + πn

tan (x) = - 1 : x = \frac{3 π}{4} + πn

tan (x) = - 1

General solutions for

tan (x) = - 1

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}

x = \frac{3 π}{4} + πn

x = \frac{3 π}{4} + πn

Combine all the solutions

x = \frac{π}{4} + πn, x = \frac{3 π}{4} + πn

x, tan^{2} (x) = 2

x, \frac{p}{q} = sin^{2} (x)

cos (2 x) = 1 - 2 (\frac{3}{5})^{2}

sin (θ + \frac{π}{3}) = \frac{1}{2}

4 sin^{2} (x) = 5 - 4 cos (x), 0 \leq x < 2 π

What is the general solution for solvefor x,tan^2(x)=1 ?
The general solution for solvefor x,tan^2(x)=1 is x= pi/4+pin,x=(3pi)/4+pin