Question

2cos^2 + cos2x = 2

Answer 1

2cos²x+cos2x=2

\(\Rightarrow\)2cos²x+2cos²x-1=2

\(\Rightarrow\)4cos²x=3

\(\Rightarrow\)cos²x=\(\dfrac{3}{4}\)

\(\Rightarrow\dfrac{1+cos2x}{2}=\dfrac{3}{4}\)

\(\Rightarrow4+4cos2x=6\)

\(\Rightarrow4cos2x=2\)

\(\Rightarrow cos2x=\dfrac{1}{2}\)

\(\Rightarrow cos2x=cos\dfrac{\Pi}{3}\)

\(\Rightarrow2x=\pm\dfrac{\Pi}{3}+k2\Pi\)

\(\Rightarrow x=\pm\dfrac{\Pi}{6}+k\Pi\)

Answer 2

\(PT\Leftrightarrow2\cos^2x-1+\cos2x=2-1=1\)\

\(\Leftrightarrow\cos2x+\cos2x=2\cos2x=1\)

\(\Leftrightarrow\cos2x=\dfrac{1}{2}\)

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

\(\Leftrightarrow x=\pm\dfrac{\pi}{6}+k\pi\left(k\in Z\right)\)

Vậy ....