Hỏi đáp Lớp 11

PT đã cho
\(\Leftrightarrow (sinx+sin3x)-4cosx=0\Leftrightarrow 2sin2xcosx-4cosx=0\)
\(\Leftrightarrow 2cosx(sin2x-2)=0\)
\(\Leftrightarrow \bigg \lbrack\begin{matrix} cosx=0\\ sin2x=2\ \ (vn) \end{matrix}\Leftrightarrow x=\frac{\pi }{2}+k\pi,k\in Z\)
KL.PT có nghiệm \(x=\frac{\pi }{2}+k\pi,k\in Z\)

Điều kiện \(sin2x\neq 0\). Ta có \(\frac{sin5x}{sinx}+\frac{cos5x}{cosx}=\frac{2sin6x}{sin2x}\)
\(=\frac{2(3sin2x-4sin^32x)}{sin2x}=6-8sin^32x=4cos4x+2\)
Ta có \(2(sin^4x+cos^4x)=2(1-2sin^2x.cos^2x)=\frac{3+cos4x}{2}\)
Phương trình đã cho tương đương với \(cosx4x=-1\Leftrightarrow x=\frac{\pi }{4}+k\frac{\pi }{2}\) (thỏa điều kiện)

\(cosx+2cosx^2\frac{x}{3}-3\Leftrightarrow 4cos^3\frac{x}{3}-3cos\frac{x}{3}+2cos^2\frac{x}{3}-3=0\)

\(\Leftrightarrow (cos\frac{x}{3}-1)(4cos^2\frac{x}{3}+6cos\frac{x}{3}+3)=0\)

\(\Leftrightarrow cos\frac{x}{3}=1\Leftrightarrow \frac{x}{3}=k2\pi \Leftrightarrow x=6k\pi , k\in Z\)

\(2sin^2x+\sqrt{3}sin2x-2=0\Leftrightarrow \sqrt{3}sin2x-cos2x=1\Leftrightarrow \frac{\sqrt{3}}{2}sin2x-\frac{1}{2}cos2x\)
\(\Leftrightarrow sin(2x-\frac{\pi }{6})=sin\frac{\pi }{6}\Leftrightarrow \bigg \lbrack \begin{matrix} x=\frac{\pi }{6}+k\pi \\ x=\frac{\pi }{6}+k\pi \end{matrix}\)

a)
\(PT\Leftrightarrow sin(2x+\frac{\pi}{4})=sin(\frac{\pi}{2}-x)\)
\(\Leftrightarrow \Bigg \lbrack \begin{matrix} 2x+\frac{\pi}{4}=\frac{\pi}{2}-x+k2\pi\\ \\ 2x+\frac{\pi}{4}=\frac{\pi}{2}+x+k2\pi \end{matrix}\)
\(\Leftrightarrow \Bigg \lbrack \begin{matrix} 3x=\frac{\pi}{4}+k2\pi\\ \\ x=\frac{\pi}{4}+k2\pi \end{matrix}\Leftrightarrow \Bigg \lbrack \begin{matrix} x=\frac{\pi}{12}+\frac{k2\pi}{3}\\ \\ x=\frac{\pi}{4}+k2\pi \end{matrix}\)
b)
\(PT\Leftrightarrow \left\{\begin{matrix} 3x=x+k\pi\\ x\neq l\pi \end{matrix}\right .\Leftrightarrow \left\{\begin{matrix} x=\frac{k\pi}{2}\\ x\neq l\pi \end{matrix}\right.\)
\(x=\frac{k\pi}{2}\) là nghiệm khi \(\frac{k\pi}{2}\neq l\pi\Leftrightarrow k\neq 2l\)
\(\Rightarrow\) k lẻ hay k = 1 + 2n
Vậy nghiệm \(x=\frac{\pi}{2}+n\pi \ (n\in Z)\)