cosx-cos2x+2=0 Giúp em với ạ

2 bình luận về “cosx-cos2x+2=0 Giúp em với ạ”

1. cos x – cos 2x + 2 = 0

   <=> cos x – (2 cos^2 x – 1) + 2 = 0

   <=> cos x – 2 cos^2 x + 1 + 2 = 0

   <=> cos x – 2 cos^2 x + 3 = 0

   <=> -2 cos^2 x + cos x + 3 = 0

   <=> 2 cos^2 x – cos x – 3 = 0

   <=> (cos x + 1) (2 cos x – 3) = 0

   <=>$[\begin{array}{l}\text{cos}x+1=0\\ 2\text{cos}x–3=0\end{array}$ 

   <=>$[\begin{array}{l}x=\pi +2n\pi \\ \text{cos}x={\displaystyle \frac{3}{2}}(\mathrm{L}\mathrm{o}\mathrm{ạ}\mathrm{i})\end{array}$ 

   <=> x = \pi + 2n\pi (n \in Z)

   Trả lời
2. cosx-cos2x+2=0

   <=>cosx-(2cos^2x-1)+2=0

   <=>cosx-2cos^2x+1+2=0

   <=>-2cos^2x+cosx+3=0   

   <=>$[\begin{array}{l}cosx={\displaystyle \frac{3}{2}}(loại)\\ cosx=-1\end{array}$ 

   =>cosx=-1

   =>x=\pi+k2\pi(k∈Z)  

   Trả lời