Điều kiện: $\cos x\ne0\Leftrightarrow x\ne\frac{\pi}{2}+k\pi,k\in\mathbb{Z}$Đặt: $\tan^2x=t,(t\ge0)$ ta có: $3(2-t)^4+4t^3=7$$\Leftrightarrow 3t^4-20t^3+72t^2-96t+41=0$$\Leftrightarrow (t-1)^2(3t^2-14t+41)=0$$\Leftrightarrow t=1$Suy ra: $\tan^2x=1\Leftrightarrow x=\pm\frac{\pi}{4}+k\pi,k\in\mathbb{Z}$, thỏa mãn...
Điều kiện: $\cos x\ne0\Leftrightarrow x\ne\frac{\pi}{2}+k\pi,k\in\mathbb{Z}$Đặt: $\tan^2x=t,(t\ge0)$ ta có: $3(2-t)^4+4t^3=7$$\Leftrightarrow 3t^4-20t^3+72t^2-96t+41=0$$\Leftrightarrow (t-1)^2(3t^2-14t+41)=0$$\Leftrightarrow t=1$Suy ra: $\tan^2x=1\Leftrightarrow x=\pm\frac{\pi}{4}+k\pi,k\in\mathbb{Z}$, thỏa mãn.
Điều kiện: $\cos x\ne0\Leftrightarrow x\ne\frac{\pi}{2}+k\pi,k\in\mathbb{Z}$Đặt: $\tan^2x=t,(t\ge0)$ ta có: $3(2-t)^4+4t^3=7$$\Leftrightarrow 3t^4-20t^3+72t^2-96t+41=0$$\Leftrightarrow (t-1)^2(3t^2-14t+41)=0$$\Leftrightarrow t=1$Suy ra: $\tan^2x=1\Leftrightarrow x=\pm\frac{\pi}{4}+k\pi,k\in\mathbb{Z}$, thỏa mãn.
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