\(A=cos^2x+cos^2\left(\dfrac{\pi}{3}+x\right)+cos^2\left(\dfrac{\pi}{3}-x\right)\)
\(\Leftrightarrow A=\dfrac{1}{2}\left(1+cos2x\right)+\dfrac{1}{2}\left[1+cos\left(\dfrac{2\pi}{3}+2x\right)\right]+\dfrac{1}{2}\left[1+cos\left(\dfrac{2\pi}{3}-2x\right)\right]\)
\(\Leftrightarrow A=\dfrac{3}{2}+\dfrac{1}{2}\left[cos2x+cos\left(\dfrac{2\pi}{3}+2x\right)+cos\left(\dfrac{2\pi}{3}-2x\right)\right]\)
mà ta có công thức \(cos\left(a-b\right)+cos\left(a+b\right)=2cosa.cosb\)
\(\Leftrightarrow A=\dfrac{3}{2}+\dfrac{1}{2}\left[cos2x+2cos\dfrac{2\pi}{3}.cos2x\right]\)
\(\Leftrightarrow A=\dfrac{3}{2}+\dfrac{1}{2}\left[cos2x+2cos\left(\pi-\dfrac{\pi}{3}\right).cos2x\right]\)
\(\Leftrightarrow A=\dfrac{3}{2}+\dfrac{1}{2}\left[cos2x-2cos\dfrac{\pi}{3}.cos2x\right]\)
\(\Leftrightarrow A=\dfrac{3}{2}+\dfrac{1}{2}\left[cos2x-2.\dfrac{1}{2}.cos2x\right]\)
\(\Leftrightarrow A=\dfrac{3}{2}\)
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