Question

Solve the following equation: -sin^2x=2cosx-2

Answer 1

$\bf sin^2(\theta)+cos^2(\theta)=1\implies sin^2(\theta)=1-cos^2(\theta)\\\\ -------------------------------\\\\\\-sin^2(x)=2cos(x)-2\implies -[1-cos^2(x)]=2cos(x)-2 \\\\\\ cos^2(x)-1=2cos(x)-2\implies \stackrel{ax^2+bx+c~form}{cos^2(x)-2cos(x)+1}=0 \\\\\\\ [cos(x)-1][cos(x)-1]=0\implies cos(x)=1\implies \measuredangle x= \begin{cases} 0\\ 2\pi \end{cases}$

Answer 2

$-\sin^2x=2\cos x-2\ \ |\text{use:}\sin^2\alpha+\cos^2\alpha=1\to \sin^2\alpha=1-\cos^2\alpha\\\\-(1-\cos^2x)=2\cos x-2\\\\-1+\cos^2x=2\cos x-2\ \ \ |-2\cos x\\\\\cos^2x-2\cos x-1=-2\ \ \ \ |+2\\\\\cos^2x-2\cos x+1=0\\\\\cos^2x-\cos x-\cos x+1=0\\\\\cos x(\cos x-1)-1(\cos x-1)=0\\\\(\cos x-1)(\cos x-1)=0\\\\(\cos x-1)^2=0\iff\cos x-1=0\ \ \ |+1\\\\\cos x=1\\\\\boxed{x=2k\pi},\ k\in\mathbb{Z}$