Question

\(2\cos\left(x+\frac{\pi}{6}\right)+3\cos\left(x+\frac{\pi}{3}\right)=2\)

Answer 1

\(2\left(\frac{\sqrt{3}}{2}cosx-\frac{1}{2}sinx\right)+3\left(\frac{1}{2}cosx-\frac{\sqrt{3}}{2}sinx\right)=2\)

\(\Leftrightarrow\frac{2\sqrt{3}+3}{2}cosx-\frac{2+3\sqrt{3}}{2}sinx=2\)

\(\Leftrightarrow\frac{2\sqrt{3}+3}{2\sqrt{13+6\sqrt{3}}}cosx-\frac{2+3\sqrt{3}}{2\sqrt{13+6\sqrt{3}}}sinx=\frac{2}{\sqrt{13+6\sqrt{3}}}\)

Đặt \(\frac{2\sqrt{3}+3}{2\sqrt{13+6\sqrt{3}}}=cosa\)

\(\Rightarrow cosx.cosa-sinx.sina=\frac{2}{\sqrt{13+6\sqrt{3}}}\)

\(\Leftrightarrow cos\left(x-a\right)=\frac{2}{\sqrt{13+6\sqrt{3}}}\)

\(\Leftrightarrow x=a\pm arccos\left(\frac{2}{\sqrt{13+6\sqrt{3}}}\right)+k2\pi\)