Hỏi đáp Lớp 11

Phương trình đã cho tương đương với:
\(\Leftrightarrow -2 sin 2x sinx = 2 sin 2x +sin x +1\)
\(\Leftrightarrow (sin x +1)( 2 sin 2x +1)=0\Leftrightarrow \bigg \lbrack\begin{matrix} sin x=- 1\\ sin2x=-\frac{1}{2} \end{matrix}\)

+ \(sinx=-1\Leftrightarrow z=-\frac{\pi }{2}+k2\pi\)
+ \(sin2x=-\frac{1}{2}\Leftrightarrow \bigg \lbrack\begin{matrix} x=-\frac{\pi }{12}+k\pi \\ x=\frac{7\pi }{12}+k\pi \\ \end{matrix}\)

\(PT\Leftrightarrow 1-sin^2x+3sinx=2\Leftrightarrow 2sin^2x-3sinx+1=0\)
\(\Leftrightarrow sinx=1\) hoặc \(sinx=\frac{1}{2}\)
* \(sinx=1\Leftrightarrow x=\frac{\pi}{2}+k2\pi\)
* \(sinx=\frac{1}{2}=\frac{\pi}{6}\Leftrightarrow \bigg \lbrack\begin{matrix} x=\frac{\pi}{6}+k2\pi\\ x=\frac{5\pi}{6}+k2\pi \end{matrix}\)

\(2cos^3x + cos 4x+ cos 2x= 0\)
\(\Leftrightarrow 2cos^3x+ 2cos3x .cosx= 0\)
\(\Leftrightarrow cosx(cos^2x+cos3x)=0\)
\(\Leftrightarrow cosx(cos^2x+4cos^3x-3cosx)=0\)
\(\Leftrightarrow \bigg \lbrack\begin{matrix} cosx=\frac{3}{4}\\ cosx=-1\\ cosx=0 \end{matrix}\)
\(\Leftrightarrow \bigg \lbrack\begin{matrix} x=\pm arccos\left ( \frac{3}{4} \right )+k2\pi \\ x=\pi +k2\pi \ \ \ \ \ \ \ \ \ \ \ \ k\in Z \\ x=\frac{\pi }{2}+k\pi \end{matrix}\)
Vậy phương trình có nghiệm: \(x=\pm arccos\left ( \frac{3}{4} \right )+k2\pi,x=\pi +k2\pi, x=\frac{\pi }{2}+k\pi , k\in Z\)

\(2cos5x.cos3x+sinx=cos8x \ \ (1)\)
\((1)\Leftrightarrow cos8x+cos2x+sinx=cos8x\)
\(\Leftrightarrow 2sin^2x-sinx-1=0\Leftrightarrow sinx=1,sinx=-\frac{1}{2}\)
+ \(sinx=1\Leftrightarrow x=\frac{\pi }{2}+k2\pi , k\in Z\)
+ \(sinx=-\frac{1}{2}\Leftrightarrow x=-\frac{\pi }{6}+k2\pi , x=\frac{7\pi }{6}+k2\pi, k\in Z\)

\(\sqrt{3}\sin 2x-\cos 2x-3 \cos (\frac{\pi}{2}-x)-\sqrt{3}\cos x+2=0\)

\(\Leftrightarrow \sqrt{3}\sin 2x-\cos 2x-3\sin x-\sqrt{3}\cos x+2=0\)

\(\Leftrightarrow 2\sqrt{3}\sin x\cos x+2\sin ^{2}x-3\sin x-\sqrt{3}\cos x+1=0\)

\(\Leftrightarrow \sqrt{3}\cos x(2\sin x-1)+(2\sin x-1)(\sin x-1)=0\)

\(\Leftrightarrow (2\sin x-1)(\sin x+\sqrt{3}\cos x-1)=0\)

\(\Leftrightarrow \bigg \lbrack\begin{matrix} \sin x=\frac{1}{2}\\\sin x+\sqrt{3}\cos x=1 \end{matrix}\Leftrightarrow \Bigg \lbrack\begin{matrix} x=\frac{\pi}{6}+k2\pi\\x=\frac{5\pi}{6}+k2\pi \\\sin (x+\frac{\pi}{3})=\frac{1}{2} \end{matrix}\Leftrightarrow \Bigg \lbrack\begin{matrix} x=\frac{\pi}{6}+k2\pi\\x=\frac{5\pi}{6}+k2\pi \\x=-\frac{\pi}{6}+k2\pi \\x=\frac{\pi}{2}+k2\pi \end{matrix}\; (k\in Z)\)

Vậy PT có 4 họ nghiệm \(x=\pm \frac{\pi}{6}+k2\pi;\: \frac{5\pi}{6}+k2\pi;\: \frac{\pi}{2}+k2\pi,\; (k\in Z)\)

\(cos2x + (1 + 2cosx).(sinx - cosx) = 0\)

\((sinx- cosx) . (cosx- sinx+1)-0\)
\(\Leftrightarrow \bigg \lbrack \begin{matrix} sinx- cosx= 0\\ cosx-sinx+1=0 \end{matrix}\Leftrightarrow \bigg \lbrack \begin{matrix} \sqrt{2}sin\left ( x-\frac{\pi }{4} \right )=0\\ \sqrt{2}sin\left ( x-\frac{\pi }{4} \right )=1 \end{matrix}\)
\(\Leftrightarrow \bigg \lbrack \begin{matrix} x=\frac{\pi }{4}+k\pi \\ x=\frac{\pi }{2}+k2\pi\\ x=\pi +k2\pi \end{matrix}, \ \ k\in R\)

Vậy pt đã cho có nghiệm \(x=\frac{\pi }{4}+k\pi ;x=\frac{\pi }{2}+k2\pi;x=\pi +k2\pi,(k\in R)\)

Ta có: \((\sin x+\cos x)^{2}=1+\cos x\Leftrightarrow 1+2\sin x\cos x=1+\cos x\)

\(\Leftrightarrow \cos x(2\sin x-1)=0\)

\(\Leftrightarrow \bigg \lbrack\begin{matrix} \cos x=0\\\sin x=\frac{1}{2} \end{matrix}\)

\(\Leftrightarrow \bigg \lbrack\begin{matrix} x=\frac{\pi}{2}+k\pi\\x=\frac{\pi}{6}+k2\pi \\x=\frac{5\pi}{6}+k2\pi \end{matrix}\)

sin2x – cosx + sinx = 1 (1)
\((1)\Leftrightarrow (sinx - cosx)(1 + sinx - cosx ) = 0\)
\(\Leftrightarrow \bigg \lbrack\begin{matrix} sinx-cosx=0\\ 1+sinx-cosx=0 \end{matrix}\Leftrightarrow \bigg \lbrack\begin{matrix} x=\frac{\pi }{4}+k\pi \\ x=2k\pi ; \ x = \frac{3\pi }{2}+2k\pi \end{matrix}(k\in Z)\)

\(cos2x+cos^2x-sinx=0\Leftrightarrow -3sin^2-sinx+4=0\Leftrightarrow sinx=1\)
\(sinx=1\Leftrightarrow x=\frac{\pi }{2}+k2\pi ,(k\in Z)\)

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