Toán Sinx – Cotπ/6Cosx=2 B, 2cos^2x -3√2cosx+2=0 13/09/2021 By Mackenzie Sinx – Cotπ/6Cosx=2 B, 2cos^2x -3√2cosx+2=0
b) $2\cos^2x – 3\sqrt{2} \cos x + 2 = 0$ $<->(\cos x – \sqrt{2})(\cos x – \dfrac{\sqrt{2}}{2}) = 0$ Vậy $\cos x = \sqrt{2}$ (loại) hoặc $\cos x = \dfrac{\sqrt{2}}{2}$ Vậy $x = \pm \dfrac{\pi}{4} + 2k\pi$. Trả lời
b)
$2\cos^2x – 3\sqrt{2} \cos x + 2 = 0$
$<->(\cos x – \sqrt{2})(\cos x – \dfrac{\sqrt{2}}{2}) = 0$
Vậy $\cos x = \sqrt{2}$ (loại) hoặc $\cos x = \dfrac{\sqrt{2}}{2}$
Vậy $x = \pm \dfrac{\pi}{4} + 2k\pi$.