Question

\(\cos\left(2x+\dfrac{\pi}{6}\right)=0\)

Answer 1

\(cos\left(2x+\dfrac{\pi}{6}\right)=0\)

\(\Leftrightarrow2x+\dfrac{\pi}{6}=\dfrac{\pi}{2}+k\pi\)

\(\Leftrightarrow2x=\dfrac{\pi}{2}-\dfrac{\pi}{6}+k\pi\)

\(\Leftrightarrow x=\dfrac{\pi}{6}+\dfrac{k\pi}{2}\left(k\in Z\right)\)

Answer 2

cos(2x + pi/6) = 0
<=> 2x + pi/6 = pi/2 + k.pi
<=> 2x = pi/3 + kpi
<=> x = pi/6 + k.pi/2 , k thuộc Z
vậy S = { pi/6 + k.pi/2 | k thuộc Z }

Answer 3

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