Question

Giải phương trình:

a,\(3Cos2x-4Sin2x=1\)

Answer 1

Pt\(\Leftrightarrow3\left(cos^2x-sin^2x\right)-8.sinx.cosx=sin^2x+cos^2x\)

\(\Leftrightarrow2cos^2x-8sinx.cosx-4sin^2x=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cosx=\left(2+\sqrt{6}\right)sinx\\cosx=\left(2-\sqrt{6}\right)sinx\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}tanx=\dfrac{1}{2+\sqrt{6}}\\tanx=\dfrac{1}{2-\sqrt{6}}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=arc.tan\left(\dfrac{1}{2+\sqrt{6}}\right)+k\pi\\x=arc.tan\left(\dfrac{1}{2-\sqrt{6}}\right)+k\pi\end{matrix}\right.\), k nguyên

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