Giúp mình với ạ, mình cảm ơn ạ ><

Đáp án:

a) \(4\sqrt 3 \left( {cm} \right)\)

b) \( \pm \dfrac{\pi }{2}\left( {rad} \right)\)

Giải thích các bước giải:

a) Ta có: 

\(\begin{array}{l}
{x_1} = A\cos \dfrac{\pi }{6} = 4\sqrt 3 \left( {cm} \right)\\
{x_2} = A\cos \dfrac{\pi }{2} = 0\left( {cm} \right)\\
{x_3} = A\cos \dfrac{{2\pi }}{3} =  - 4\left( {cm} \right)\\
{x_4} = A\cos \left( { - \dfrac{\pi }{3}} \right) = 4\left( {cm} \right)\\
{x_5} = A\cos \dfrac{{7\pi }}{6} =  - 4\sqrt 3 \left( {cm} \right)
\end{array}\)

b) Ta có: 

\(\begin{array}{l}
\cos {\varphi _1} = \dfrac{{{x_1}}}{A} = 0 \Rightarrow {\varphi _1} =  \pm \dfrac{\pi }{2}\left( {rad} \right)\\
\cos {\varphi _2} = \dfrac{{{x_2}}}{A} = \dfrac{4}{8} = \dfrac{1}{2} \Rightarrow {\varphi _2} =  \pm \dfrac{\pi }{3}\left( {rad} \right)\\
\cos {\varphi _3} = \dfrac{{{x_3}}}{A} = \dfrac{{ - 4\sqrt 2 }}{8} =  - \dfrac{{\sqrt 2 }}{2} \Rightarrow {\varphi _3} =  \pm \dfrac{{3\pi }}{4}\left( {rad} \right)\\
\cos {\varphi _4} = \dfrac{{{x_4}}}{A} =  - \dfrac{8}{8} = 1 \Rightarrow {\varphi _4} = \pi \left( {rad} \right)
\end{array}\)

\(a)\begin{cases} {x_1} = A\cos \dfrac{\pi }{6}=8\cos \dfrac{\pi }{6} = 4\sqrt 3 \left( {cm} \right)\\ {x_2} = A\cos \dfrac{\pi }{2} =8\cos \dfrac{\pi }{2}= 0\left( {cm} \right)\\ {x_3} = A\cos \dfrac{{2\pi }}{3} =8\cos \dfrac{{2\pi }}{3}= - 4\left( {cm} \right)\\ {x_4} = A\cos \left( { - \dfrac{\pi }{3}} \right) =8\cos \left( { - \dfrac{\pi }{3}} \right) = 4\left( {cm} \right)\\ {x_5} = A\cos \dfrac{{7\pi }}{6}=8\cos \dfrac{{7\pi }}{6} = - 4\sqrt 3 \left( {cm} \right) \end{cases}\\b)\begin{cases}\cos {\varphi _1} = \dfrac{{{x_1}}}{A} = 0 \Rightarrow {\varphi _1} = \pm \dfrac{\pi }{2}\left( {rad} \right)\\ \cos {\varphi _2} = \dfrac{{{x_2}}}{A} = \dfrac{4}{8} = \dfrac{1}{2} \Rightarrow {\varphi _2} = \pm \dfrac{\pi }{3}\left( {rad} \right)\\ \cos {\varphi _3} = \dfrac{{{x_3}}}{A} = \dfrac{{ - 4\sqrt 2 }}{8} = - \dfrac{{\sqrt 2 }}{2} \Rightarrow {\varphi _3} = \pm \dfrac{{3\pi }}{4}\left( {rad} \right)\\ \cos {\varphi _4} = \dfrac{{{x_4}}}{A} = - \dfrac{8}{8} = 1 \Rightarrow {\varphi _4} = \pi \left( {rad} \right) \end{cases}\)