How do you solve $ y = - \dfrac{1}{2}\cos \left( {\dfrac{\pi }{3}} \right)x? $ ?

Hint: In order to determine the period and amplitude of the above trigonometric. Compare the cosine function with the standard cosine function i.e. $ y = A\cos \left( {Bx - C} \right) + D $ to find the value of $ A,B,C,D $ . Amplitude is equal to the modulus of A , And period of the function is equal to ratio of $ 2\pi $ and modulus of $ B $

Complete step-by-step answer:
We are given a trigonometric function $ y = - \dfrac{1}{2}\cos \left( {\dfrac{\pi }{3}} \right)x $
Comparing this equation with the standard cosine function $ y = A\cos \left( {Bx - C} \right) + D $ we get
 $ A = - \dfrac{1}{2},B = \dfrac{\pi }{3},C = D = 0 $
Amplitude is equal to the modulus of A i.e.
Amplitude $ = \left| A \right| = \left| { - \dfrac{1}{2}} \right| = \dfrac{1}{2} $
And period of the function is equal to ratio of $ 2\pi $ and modulus of $ B $
Period = $ \dfrac{{2\pi }}{{\left| B \right|}} = \dfrac{{2\pi }}{{\dfrac{\pi }{3}}} = 6 $
Therefore the amplitude $ A $ and period of the cosine function
 $ y = - \dfrac{1}{2}\cos \left( {\dfrac{\pi }{3}} \right)x $ is equal to $ \dfrac{1}{2} $ and $ 6 $ respectively.
So, the correct answer is “ $ y = - \dfrac{1}{2}\cos \left( {\dfrac{\pi }{3}} \right)x $ is equal to $ \dfrac{1}{2} $ and $ 6 $ ”.

Note: 1. Periodic Function= A function $ f(x) $ is said to be a periodic function if there exists a real number T > 0 such that $ f(x + T) = f(x) $ for all x.
If T is the smallest positive real number such that $ f(x + T) = f(x) $ for all x, then T is called the fundamental period of $ f(x) $ .
Since $ \sin \,(2n\pi + \theta ) = \sin \theta $ for all values of $ \theta $ and n $ \in $ N.
2. Even Function – A function $ f(x) $ is said to be an even function ,if $ f( - x) = f(x) $ for all x in its domain.
Odd Function – A function $ f(x) $ is said to be an even function ,if $ f( - x) = - f(x) $ for all x in its domain.
We know that $ \sin ( - \theta ) = - \sin \theta .\cos ( - \theta ) = \cos \theta \,and\,\tan ( - \theta ) = - \tan \theta $
Therefore, $ \sin \theta $ and $ \tan \theta $ and their reciprocals, $ \cos ec\theta $ and $ \cot \theta $ are odd functions whereas \[\cos \theta \] and its reciprocal \[\sec \theta \] are even functions.