\Leftrightarrow1, 3cosx+ 2|sinx| =2
\Leftrightarrow 2|sinx| = 2-3cosx ( dk: cosx \leq 2/3)
\Leftrightarrow (2|sinx|)^2 = (2-3cosx)^2
\Leftrightarrow 4 sin^2[x] = 4 - 12 cos x + 9cos^2[x]
\Leftrightarrow (4 sin^2[x] + 4 cos^2[x] )-4+12cosx -13cos^2[x] =0
\Leftrightarrow 4 -4 +12cosx -13cos^2[x] = 0
\Leftrightarrow 13 cos^2[x] + 12cosx = 0 dat cosx =t suy ra pt tuong duong voi 13t^2 - 12t =0 ( dk |t|\leq1)
\Leftrightarrow hoac t= 12/13 hoac t= 0 voi t=12/13(ko thoa man dk : t\leq 2/3) voi t= o \Leftrightarrow cosx = 0 \Leftrightarrow x = pi/2 +k2[pi] >>> O bai vua nay toi chi thu ban thoi, trung muu roi nha , ban bao la khong hieu bai ma sao ban nhan xet tui ky luong the, thuc ra ban nhan xet van con thieu day, vi bai nay phai can co dieu kien de sau khi giai ta se doi chieu voi dkien de loai nghiem va lay nghiem thich hop,du saoo cung thanks ban nha nha nha!!>>>