Hỏi đáp Lớp 11

\(2cos2x+8sinx-5=0\)
\(\Leftrightarrow -4sin^2x+8sinx-3=0\)
\(\Leftrightarrow \bigg \lbrack\begin{matrix} sinx=\frac{3}{2}\\ sinx=\frac{1}{2} \end{matrix}\)
\(\Leftrightarrow x=\frac{\pi}{6}+k2\pi,x=\frac{5\pi}{6}+k2\pi\)

Ta có \((1)\Leftrightarrow 2sin2xcos\frac{\pi}{3}+2cosxsin\frac{\pi}{3}-\sqrt{3}cos2x=-2\)

\(\Leftrightarrow sin2x+\sqrt{3}cos2x-\sqrt{3}cos2x=-2\Leftrightarrow sin2x=-2\)   (2)
Do \(\left | sin2x \right |\leq 1\) nên phương trình (2) vô nghiệm

\(sin3x-sin2x+sinx=0\Leftrightarrow (sin3x+sinx)-sin2x=0\)
\(\Leftrightarrow 2sin2xcos2x-sin2x=0\Leftrightarrow sin2x(2cosx-1)=0\)
\(\Leftrightarrow \bigg \lbrack\begin{matrix} sin2x=0\\ cosx=\frac{1}{2} \end{matrix}\Leftrightarrow \bigg \lbrack\begin{matrix} 2x=k\pi\\ x=\frac{\pm \pi}{3}+k2\pi \end{matrix}\Leftrightarrow \bigg \lbrack\begin{matrix} x=\frac{k\pi}{2}\\ x= \frac{\pm \pi}{3}+k2\pi \end{matrix}\)
Vậy phương trình có nghiệm \(\bigg \lbrack\begin{matrix} x=\frac{k\pi}{2}\\ x= \frac{\pm \pi}{3}+k2\pi \end{matrix}(k\in Z)\)

\(Pt \Leftrightarrow \bigg \lbrack\begin{matrix} sinx=0\\ cosx=\frac{\sqrt{2}}{2} \end{matrix}\)

\(\Leftrightarrow \Bigg \lbrack\begin{matrix} x=k\pi \ \ \ \ \ \ \ \\ x=\frac{\pi}{4}+k2\pi \ \ \ \\ x=-\frac{\pi}{4}+k2\pi \end{matrix}\)

\(PT\Leftrightarrow sin2x+1+cos2x-4cosx=0\)
\(\Leftrightarrow 2sinxcosx+2cos^2x-4cosx=0\)
\(\Leftrightarrow cosx(sinx+cosx-2)=0\)
\(\Leftrightarrow \bigg \lbrack \begin{matrix} cosx=0\\ sinx+cosx=2 \ (VN\ do 1^2+1^2<2^2 ) \end{matrix}\Leftrightarrow x=\frac{\pi}{2}+k\pi\)
Vậy nghiệm của phương trình đã cho là: \(x=\frac{\pi}{2}+k\pi\)

\(\sin 3x+\cos 2x=1+2\sin x\cos 2x\Leftrightarrow \sin 3x+\cos 2x=1-\sin x+\sin 3x\)

\(\Leftrightarrow \cos 2x=1-\sin x\)

\(\Leftrightarrow 1-2\sin ^{2}x=1-\sin x\Leftrightarrow \bigg \lbrack\begin{matrix} \sin x=0\\ \sin x=\frac{1}{2} \end{matrix}\Leftrightarrow \Bigg \lbrack\begin{matrix} x=k\pi\\ x=\frac{\pi}{6}+k2\pi \\ x=\frac{5\pi}{6}+k2\pi \end{matrix}\)

\(sin3x-sinx+cos2x=1\Leftrightarrow 2cos2xsinx-2sin^2x=0\)
\(\Leftrightarrow \bigg \lbrack\begin{matrix} sinx=0\\ cos2x=sinx \end{matrix}\)
\(+sinx=0\Leftrightarrow x=k\pi,k\in Z\)
\(+cos2x=sinx\Leftrightarrow cos2x=cos\left ( \frac{\pi}{2} -x\right )\Leftrightarrow \bigg \lbrack\begin{matrix} x=\frac{\pi}{6}+k\frac{2\pi}{3}\\ x=-\frac{\pi}{2}+k2\pi \end{matrix}(k\in Z)\)

\(cos2x+5sinx-3=0\Leftrightarrow -2sin^2x+5sinx-2=0\)
\(\Leftrightarrow (sinx-2)(2sinx-1)=0 \ (Do \ sinx-2<0, \forall x)\)
\(\Leftrightarrow 2sinx-1=0\Leftrightarrow \bigg \lbrack\begin{matrix} x=\frac{\pi}{6}+k2\pi\\ x=\frac{5\pi}{6}+k2\pi \end{matrix} \ (k\in Z)\)

\(cos^23x+sin^22x=1\)
\(\Leftrightarrow \frac{cos6x+1}{2}+\frac{1-cos4x}{2}=1\Leftrightarrow cos6x=cos4x\)
\(\Leftrightarrow \bigg \lbrack \begin{matrix} 6x=4x+k2\pi\\ 6x=-4x+k2\pi \end{matrix}\Leftrightarrow \bigg \lbrack\begin{matrix} x=k\pi\\ x=\frac{k\pi}{5} \end{matrix}\Leftrightarrow x=\frac{k\pi}{5}\)

\(\Leftrightarrow cos2x-\sqrt{3}sin2x=\sqrt{3}sinx+cosx\)
\(\Leftrightarrow \frac{1}{2}cos2x-\frac{\sqrt{3}}{2}sin2x=\frac{\sqrt{3}}{2}sinx+\frac{1}{2}cosx\)
\(\Leftrightarrow cos(2x+\frac{\pi}{3})=cos(x-\frac{\pi}{3})\Leftrightarrow \Bigg \lbrack \begin{matrix} 2x+\frac{\pi}{3}=x-\frac{\pi}{3}+k2\pi\\ \\ 2x+\frac{\pi}{3}=-x+\frac{\pi}{3}+k2\pi \end{matrix}\Leftrightarrow \Bigg \lbrack\begin{matrix} x=-\frac{2\pi}{3}+k2\pi\\ \\ x=k\frac{2\pi}{3} \end{matrix}\), \(k\in Z\)