The general solution of the equation sin 2 θ sec &the...

θ = nπ + (-1) n+1 · \(\frac{\pi}{3}\)

θ = nπ + (-1) n+1 · \(\frac{\pi}{6}\)

The general solution of the equation sin²θ sec θ + √3 tan θ = 0 is:

θ = nπ + (-1)ⁿ⁺¹ · \(\frac{\pi}{3}\)

θ = nπ

θ = nπ + (-1)ⁿ⁺¹ · \(\frac{\pi}{6}\)

θ = \(\rm n\frac{\pi}{2}\) where n ∈ Zwhere n ∈ Z