PrecalculusEasy SharePrecalculus · Easy problemSolve 2cosx+2=02\cos x+\sqrt2=02cosx+2=0 on [0,2π)[0,2\pi)[0,2π).A43π, 5π4\dfrac{4}{3\pi},\ \dfrac{5\pi}{4}3π4, 45πB6π4, 5π4\dfrac{6\pi}{4},\ \dfrac{5\pi}{4}46π, 45πC3π4, 5π5\dfrac{3\pi}{4},\ \dfrac{5\pi}{5}43π, 55πD3π4, 5π4\dfrac{3\pi}{4},\ \dfrac{5\pi}{4}43π, 45πCheck Show answer Walk through it step by step Practice more like this