What is the general solution for sqrt(3)tan(θ-20)=tan^2(45) ?

The general solution for sqrt(3)tan(θ-20)=tan^2(45) is θ=180n+50

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sqrt(3)tan(θ-20)=tan^2(45)

3 tan (θ - 2 0^{\circ}) = tan^{2} (4 5^{\circ})

θ = 18 0^{\circ} n + 5 0^{\circ}

Solution steps

3 tan (θ - 2 0^{\circ}) = tan^{2} (4 5^{\circ})

tan (4 5^{\circ}) = 1

3 tan (θ - 2 0^{\circ}) = 1^{2}

Apply rule

1^{a} = 1

3 tan (θ - 2 0^{\circ}) = 1

Divide both sides by

3

tan (θ - 2 0^{\circ}) = \frac{3}{3}

General solutions for

tan (θ - 2 0^{\circ}) = \frac{3}{3}

θ - 2 0^{\circ} = 3 0^{\circ} + 18 0^{\circ} n

Solve

θ - 2 0^{\circ} = 3 0^{\circ} + 18 0^{\circ} n : θ = 18 0^{\circ} n + 5 0^{\circ}

θ = 18 0^{\circ} n + 5 0^{\circ}

0 = - 2 sin (x) - 4 cos (2 x)

- 0.1429 = cos (C)

sin^{2} (x) = 1 - cos (x), 0 \leq x \leq 2 π

cos (\frac{π}{4} c) = sin (2 (\frac{π}{4}))

\frac{( 1 - tanh ( 2 x ) )}{( 1 + tanh ( 2 x ) )} = 2

What is the general solution for sqrt(3)tan(θ-20)=tan^2(45) ?
The general solution for sqrt(3)tan(θ-20)=tan^2(45) is θ=180n+50