What is the general solution for 2tan^2(θ)=2 ?

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

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2tan^2(θ)=2

2 tan^{2} (θ) = 2

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

Degrees

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

Solution steps

2 tan^{2} (θ) = 2

Solve by substitution

2 tan^{2} (θ) = 2

Let:

tan (θ) = u

2 u^{2} = 2

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

2 u^{2} = 2

Divide both sides by

2

2 u^{2} = 2

Divide both sides by

2

\frac{2 u ^{2}}{2} = \frac{2}{2}

Simplify

u^{2} = 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 (θ)

tan (θ) = 1, tan (θ) = - 1

tan (θ) = 1, tan (θ) = - 1

tan (θ) = 1 : θ = \frac{π}{4} + πn

tan (θ) = 1

General solutions for

tan (θ) = 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}

θ = \frac{π}{4} + πn

θ = \frac{π}{4} + πn

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

tan (θ) = - 1

General solutions for

tan (θ) = - 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}

θ = \frac{3 π}{4} + πn

θ = \frac{3 π}{4} + πn

Combine all the solutions

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

0 = (6 cos (30 t) + 2 sin (30 t))

- 2 sin (θ) = 2

tan (2 x - \frac{π}{6}) = \frac{1}{3 ^{\frac{1}{2}}}

sin^{2} (\frac{x}{2}) - 1 = 0

69 = cos (x)

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