Question

\(\dfrac{cos2x-sinx-cosx+2}{sinx-1}=0\)

Answer 1

ĐKXĐ: sin x-1<>0

=>sin x<>1

=>\(x<>\frac{\pi}{2}+k2\pi\)

\(\frac{cos2x-\sin x-cosx+2}{\sin x-1}=0\)

=>cos2x-sin x-cosx+2=0

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

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

\(\iff2\cos^2x-\cos x+1-\sin x=0\)

\(2\cos^2 x - \cos x + 1 = 2\left(\cos x - \frac{1}{4}\right)^2 + \frac{7}{8} > 0\)

1-sin x>0(Do sin x<1)

Do đó: \(2\cos^2x-\cos x+1-\sin x>0\forall x\)

=>Phương trình vô nghiệm