Question

Giải pt

Answer 1

d, \(cosx-cos2x=sin3x\)

\(\Leftrightarrow2sin\dfrac{3x}{2}.sin\dfrac{x}{2}=2sin\dfrac{3x}{2}.cos\dfrac{3x}{2}\)

\(\Leftrightarrow2sin\dfrac{3x}{2}.\left(sin\dfrac{x}{2}-cos\dfrac{3x}{2}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sin\dfrac{3x}{2}=0\\sin\dfrac{x}{2}=cos\dfrac{3x}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}sin\dfrac{3x}{2}=0\\cos\left(\dfrac{\pi}{2}-\dfrac{x}{2}\right)=cos\dfrac{3x}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{3x}{2}=k\pi\\\dfrac{\pi}{2}-\dfrac{x}{2}=\pm\dfrac{3x}{2}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{k2\pi}{3}\\x=\dfrac{\pi}{4}-k\pi\\x=-\dfrac{\pi}{2}+k2\pi\end{matrix}\right.\)

Answer 2

Đăng tầm 2, 3 câu 1 lần thôi.

Answer 3

e, \(sinx.cosx+\sqrt{3}sin2x=0\)

\(\Leftrightarrow\dfrac{1}{2}sin2x+\sqrt{3}sin2x=0\)

\(\Leftrightarrow\left(\sqrt{3}+\dfrac{1}{2}\right)sin2x=0\)

\(\Leftrightarrow2x=k\pi\)

\(\Leftrightarrow x=\dfrac{k\pi}{2}\)

Answer 4

f, \(cosx-cos2x-1=0\)

\(\Leftrightarrow cosx-2cos^2x=0\)

\(\Leftrightarrow cosx\left(1-2cosx\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cosx=0\\cosx=\dfrac{1}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{2}+k\pi\\x=\pm\dfrac{\pi}{3}+k2\pi\end{matrix}\right.\)

Answer 5

g, \(\sqrt{3}sinx-cosx=\sqrt{2}\)

\(\Leftrightarrow\dfrac{\sqrt{3}}{2}sinx-\dfrac{1}{2}cosx=\dfrac{\sqrt{2}}{2}\)

\(\Leftrightarrow sin\left(x-\dfrac{\pi}{6}\right)=\dfrac{\sqrt{2}}{2}\)

\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{\pi}{6}=\dfrac{\pi}{4}+k2\pi\\x-\dfrac{\pi}{6}=\pi-\dfrac{\pi}{4}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5\pi}{12}+k2\pi\\x=\dfrac{11\pi}{12}+k2\pi\end{matrix}\right.\)

Answer 6

h, \(cos8x+sin4x=0\)

\(\Leftrightarrow1-2sin^24x+sin4x=0\)

\(\Leftrightarrow\left(sin4x-1\right)\left(2sin4x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sin4x=1\\sin4x=-\dfrac{1}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}4x=\dfrac{\pi}{2}+k2\pi\\4x=-\dfrac{\pi}{6}+k2\pi\\4x=\pi+\dfrac{\pi}{6}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{8}+\dfrac{k\pi}{2}\\x=-\dfrac{\pi}{24}+\dfrac{k\pi}{2}\\x=\dfrac{7\pi}{24}+\dfrac{k\pi}{2}\end{matrix}\right.\)

Answer 7

i, \(cos2x+cosx-2=0\)

\(\Leftrightarrow2cos^2x+cosx-3=0\)

\(\Leftrightarrow\left(cosx-1\right)\left(2cosx+3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cosx=1\\cosx=-\dfrac{3}{2}\left(l\right)\end{matrix}\right.\)

\(\Leftrightarrow x=k2\pi\)