Question

Dùng cung liên kết giải các phương trình sau:

a) Cos6x +3Cos2x = \(\sqrt{2}\)

b) Sin²x - Cos²3x = 0

c) Tan3x.tanx = 1

d) Sin3x.Cos5x = Sinx.Cos3x
 

Answer 1

a.

\(\Leftrightarrow4cos^32x-3cos2x+3cos2x=\sqrt{2}\)

\(\Leftrightarrow cos^32x=\dfrac{\sqrt{2}}{4}\)

\(\Leftrightarrow cos2x=\dfrac{\sqrt{2}}{2}\)

\(\Leftrightarrow2x=\pm\dfrac{\pi}{4}+k2\pi\)

\(\Leftrightarrow x=\pm\dfrac{\pi}{8}+k\pi\) (\(k\in Z\))

Answer 2

c.

ĐKXĐ: \(\left\{{}\begin{matrix}cos3x\ne0\\cosx\ne0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x\ne\dfrac{\pi}{6}+\dfrac{k\pi}{3}\\x\ne\dfrac{\pi}{2}+k\pi\end{matrix}\right.\)

\(tan3x.tanx=1\)

\(\Rightarrow tan3x=\dfrac{1}{tanx}\)

\(\Rightarrow tan3x=cotx\)

\(\Rightarrow tan3x=tan\left(\dfrac{\pi}{2}-x\right)\)

\(\Rightarrow3x=\dfrac{\pi}{2}-x+k\pi\)

\(\Rightarrow x=\dfrac{\pi}{8}+\dfrac{k\pi}{4}\)

Answer 3

b.

\(\Leftrightarrow\dfrac{1}{2}-\dfrac{1}{2}cos2x-\left(\dfrac{1}{2}+\dfrac{1}{2}cos6x\right)=0\)

\(\Leftrightarrow cos6x=-cos2x\)

\(\Leftrightarrow cos6x=cos\left(\pi-2x\right)\)

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

\(\Leftrightarrow\left[{}\begin{matrix}8x=\pi+k2\pi\\4x=-\pi+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{8}+\dfrac{k\pi}{4}\\x=-\dfrac{\pi}{4}+\dfrac{k\pi}{2}\end{matrix}\right.\)

(\(k\in Z\))

Answer 4

d.

\(\Leftrightarrow\dfrac{1}{2}sin8x-\dfrac{1}{2}sin2x=\dfrac{1}{2}sin4x-\dfrac{1}{2}sin2x\)

\(\Leftrightarrow sin8x=sin4x\)

\(\Leftrightarrow\left[{}\begin{matrix}8x=4x+k2\pi\\8x=\pi-4x+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{k\pi}{2}\\x=\dfrac{\pi}{12}+\dfrac{k\pi}{6}\end{matrix}\right.\)