4cos^2x - 2sinx + 2sin2x - 4cosx +1=0

Đáp án + Giải thích các bước giải:

Ta có:`4cos^2x - 2sinx + 2sin2x - 4cosx + 1 = 0`

`\iff (4cos^2x - 4cosx + 1) + (2sin2x - 2sinx) = 0`

`\iff (2cosx-1)^2 + 2sinx(2cosx - 1)=0`

`\iff (2cosx-1)(2cosx - 1 + 2sinx)=0`

`\iff` $\left[\begin{matrix} 2cosx - 1 = 0 \\ 2sinx + 2cosx - 1 = 0\end{matrix}\right.$

`\iff` $\left[\begin{matrix} cosx  = \frac{1}{2} \\ 2(sinx + cosx)  = 1\end{matrix}\right.$

`\iff` $\left[\begin{matrix} x = \pm \frac{\pi}{3} + 2k\pi \\ sin(x + \frac{\pi}{4})  = \frac{1}{2\sqrt{2}}\end{matrix}\right.$

`\iff` $\left[\begin{matrix} x = \pm \frac{\pi}{3} + 2k\pi \\ x + \frac{\pi}{4}  = arcsin\frac{1}{2\sqrt{2}} + k2\pi \\ x + \frac{\pi}{4}  = \pi - arcsin\frac{1}{2\sqrt{2}} + k2\pi\end{matrix}\right.$

`\iff` $\left[\begin{matrix} x = \pm \frac{\pi}{3} + 2k\pi \\ x  = arcsin\frac{1}{2\sqrt{2}} - \frac{\pi}{4} + k2\pi \\ x   = \frac{3\pi}{4} - arcsin\frac{1}{2\sqrt{2}} + k2\pi\end{matrix}\right.$

Vậy:`S = {\pm \frac{\pi}{3} + 2k\pi ; arcsin\frac{1}{2\sqrt{2}} - \frac{\pi}{4} + k2\pi ; \frac{3\pi}{4} - arcsin\frac{1}{2\sqrt{2}} + k2\pi | k \in ZZ}`