Question

giải PT sau:

\(\sin^2x+\cos x-1=0\)

Answer 1

tham khảo

 

Answer 2

\(\Leftrightarrow1-\cos^2x+\cos x-1=0\)

\(\Leftrightarrow-\cos^2x+\cos x=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\cos x=0\\\cos=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{2}+k2\pi\\x=k2\pi\end{matrix}\right.\left(k\in Z\right)\)

Answer 3

<=> 1- cos2x + cosx - 1 = 0

<=> (1-cosx)(1+cosx) + cosx-1 = 0

<=> (1-cosx) ( 1+cosx - 1)= 0

<=> (1-cosx) cosx = 0

<=> cosx = 1 hoặc cosx = 0

<=> x=0 hoặc x= 90