Question

[Nguyễn Thanh Điền]

Giải phương trình:

3sin2x + 2cos2x = 3

a) cos2x - sinx + cosx = 0

b) 2cos³x + sinx + cos2 = 0

Answer 1

[Nguyễn Lê Phước Thịnh]

a: \(\sqrt{3^2+2^2}=\sqrt{13}\)

Chia hai vế cho căn 13, ta được:

\(\dfrac{3}{\sqrt{13}}\cdot\sin2x+\dfrac{2}{\sqrt{13}}\cdot\cos2x=\dfrac{3}{\sqrt{13}}\)

Đặt \(\cos a=\dfrac{3}{\sqrt{13}}\)

Ta được phương trình: \(\sin\left(2x+a\right)=\cos a=\sin\left(\dfrac{\Pi}{2}-a\right)\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\left(\dfrac{\Pi}{2}-2a+k2\Pi\right)\\x=\dfrac{\Pi}{4}+k\Pi\end{matrix}\right.\)

b: \(\Leftrightarrow cos^2x-sin^2x+cosx-sinx=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}\cos x=\cos\left(\dfrac{\Pi}{2}-x\right)\\\sin\left(x-\dfrac{\Pi}{4}\right)=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\Pi}{2}-x+k2\Pi\\x=-\dfrac{\Pi}{2}+x+k2\Pi\\x-\dfrac{\Pi}{4}=-\dfrac{\Pi}{2}+k2\Pi\end{matrix}\right.\)

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