Question

[Anh Anh]

cos²x-√3 sin2x=1+sin²x

10cos²x-5sinxcosx+3sin²x=4

Mọi người giải chi tiết giúp em với ạ em cảm ơn

Answer 1

[Hồng Phúc]

1.

\(cos^2x-\sqrt{3}sin2x=1+sin^2x\)

\(\Leftrightarrow cos2x-\sqrt{3}sin2x=1\)

\(\Leftrightarrow\dfrac{1}{2}cos2x-\dfrac{\sqrt{3}}{2}sin2x=\dfrac{1}{2}\)

\(\Leftrightarrow cos\left(2x+\dfrac{\pi}{3}\right)=\dfrac{1}{2}\)

\(\Leftrightarrow2x+\dfrac{\pi}{3}=\pm\dfrac{\pi}{3}+k2\pi\)

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

Answer 2

[Hồng Phúc]

2.

\(10cos^2x-5sinx.cosx+3sin^2x=4\)

\(\Leftrightarrow20cos^2x-10sinx.cosx+6sin^2x=8\)

\(\Leftrightarrow20cos^2x-10-10sinx.cosx+6sin^2x-3=-5\)

\(\Leftrightarrow7cos2x-5sin2x=-5\)

\(\Leftrightarrow\sqrt{74}\left(\dfrac{7}{\sqrt{74}}cos2x-\dfrac{5}{\sqrt{74}}sin2x\right)=-5\)

\(\Leftrightarrow cos\left(2x+arccos\dfrac{7}{\sqrt{74}}\right)=-\dfrac{5}{\sqrt{74}}\)

\(\Leftrightarrow2x+arccos\dfrac{7}{\sqrt{74}}=\pm arccos\dfrac{5}{\sqrt{74}}+k2\pi\)

\(\Leftrightarrow x=-\dfrac{1}{2}arccos\dfrac{7}{\sqrt{74}}\pm\dfrac{1}{2}arccos\dfrac{5}{\sqrt{74}}+k\pi\)