Question

\(cos^2x+4sinx-4=0\\ 2cos2x+cosx=1\)

Answer 1

a: \(\Leftrightarrow1-sin^2x+4sinx-4=0\)

\(\Leftrightarrow-sin^2x+4sinx-3=0\)

\(\Leftrightarrow sinx=1\)

hay \(x=\dfrac{\Pi}{2}+k2\Pi\)

b: \(\Leftrightarrow\left(2cos2x-1\right)+cosx=0\)

\(\Leftrightarrow\left[2\left(2cos^2x-1\right)-1\right]+cosx=0\)

\(\Leftrightarrow4cos^2x+cosx-3=0\)

\(\Leftrightarrow\left(cosx+1\right)\left(4cosx-3\right)=0\)

=>cosx=3/4 hoặc cos x=-1

hay \(x=\pm arccos\left(\dfrac{3}{4}\right)+k2\Pi\) hoặc \(x=k\Pi\)