Question

Sin2x+4=8cosx+sinx

Answer 1

đề sai k bạn

Answer 2

<=> 2.sinx.cosx + 4 - 8.cosx - sinx = 0
<=> sinx.(2cosx - 1) - 4.(2cosx - 1) = 0
<=> (2cosx - 1).(sinx - 4) = 0
<=> \(\left[\begin{array}{nghiempt}2cosx-1=0\left(1\right)\\sinx-4=0\left(VN\right)\end{array}\right.\)
<=> (1) <=> cosx = \(\frac{1}{2}\)
<=> cosx = cos\(\frac{\pi}{3}\)
<=> \(\left[\begin{array}{nghiempt}x=\frac{\pi}{3}+2k\pi\\x=-\frac{\pi}{3}+2k\pi\end{array}\right.\) (k thuộc Z)