PrecalculusHard SharePrecalculus · Hard problemSolve tan2x−secx=1\tan^2 x-\sec x=1tan2x−secx=1 on [0,2π)[0,2\pi)[0,2π).A3π, π, 5π3\dfrac{3}{\pi},\ \pi,\ \dfrac{5\pi}{3}π3, π, 35πBπ3, π, 5π3\dfrac{\pi}{3},\ \pi,\ \dfrac{5\pi}{3}3π, π, 35πCπ6, π, 5π3\dfrac{\pi}{6},\ \pi,\ \dfrac{5\pi}{3}6π, π, 35πDπ3, π, 5π4\dfrac{\pi}{3},\ \pi,\ \dfrac{5\pi}{4}3π, π, 45πCheck Show answer Walk through it step by step Practice more like this