What is the general solution for tan^2(2θ)-3=0 ?

The general solution for tan^2(2θ)-3=0 is θ= pi/6+(pin)/2 ,θ= pi/3+(pin)/2

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tan^2(2θ)-3=0

tan^{2} (2 θ) - 3 = 0

θ = \frac{π}{6} + \frac{πn}{2}, θ = \frac{π}{3} + \frac{πn}{2}

Solution steps

tan^{2} (2 θ) - 3 = 0

Solve by substitution

tan (2 θ) = 3, tan (2 θ) = - 3

tan (2 θ) = 3 : θ = \frac{π}{6} + \frac{πn}{2}

tan (2 θ) = - 3 : θ = \frac{π}{3} + \frac{πn}{2}

Combine all the solutions

θ = \frac{π}{6} + \frac{πn}{2}, θ = \frac{π}{3} + \frac{πn}{2}

cos (2 x) = 5 sin^{2} (x) - cos^{2} (x)

3 sec (5 x) - 2 = 0

cot^{2} (x) sec (x) - sec (x) = 2 cot^{2} (x) - 2

5 cot^{2} (x) - 2 csc (x) + 2 = 0

\frac{sin ( 7 5 ^{\circ} )}{101} = \frac{sin ( b )}{65}

What is the general solution for tan^2(2θ)-3=0 ?
The general solution for tan^2(2θ)-3=0 is θ= pi/6+(pin)/2 ,θ= pi/3+(pin)/2