What is the general solution for tan^2(θ)= 1/3 ,0<= θ<= 2pi ?

The general solution for tan^2(θ)= 1/3 ,0<= θ<= 2pi is θ= pi/6 ,θ=(7pi)/6 ,θ=(5pi)/6 ,θ=(11pi)/6

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tan^2(θ)= 1/3 ,0<= θ<= 2pi

tan^{2} (θ) = \frac{1}{3}, 0 \leq θ \leq 2 π

θ = \frac{π}{6}, θ = \frac{7 π}{6}, θ = \frac{5 π}{6}, θ = \frac{11 π}{6}

Solution steps

tan^{2} (θ) = \frac{1}{3}, 0 \leq θ \leq 2 π

Solve by substitution

tan (θ) = \frac{1}{3}, tan (θ) = - \frac{1}{3}

tan (θ) = \frac{1}{3}, 0 \leq θ \leq 2 π : θ = \frac{π}{6}, θ = \frac{7 π}{6}

tan (θ) = - \frac{1}{3}, 0 \leq θ \leq 2 π : θ = \frac{5 π}{6}, θ = \frac{11 π}{6}

Combine all the solutions

θ = \frac{π}{6}, θ = \frac{7 π}{6}, θ = \frac{5 π}{6}, θ = \frac{11 π}{6}

4 sin (2 x) = 23

3 cos (2 θ) - 5 cos (θ) - 1 = 0

3 cos (b) + 3 = 0

csc (x) = \frac{21}{5}

3 cot (2 x) + 3 = 0

What is the general solution for tan^2(θ)= 1/3 ,0<= θ<= 2pi ?
The general solution for tan^2(θ)= 1/3 ,0<= θ<= 2pi is θ= pi/6 ,θ=(7pi)/6 ,θ=(5pi)/6 ,θ=(11pi)/6