Question

if $\cos p\theta =\cos q\theta ,p\ne q$then
A. \[\theta =2n\pi \]
B. \[\theta =\dfrac{2n\pi }{p\pm q}\]
C. \[\theta =\dfrac{2n\pi }{p+q}\]
D. None of these.

Answer 1

Hint: we will apply the theorem which states that for any of the real numbers $x,y$, $\cos x=\cos y$ implies that \[x=2n\pi \pm y\] where \[n\in Z\].
After applying the theorem we will get an equation which we will simplify and derive the general equation of $\theta $ for the given question.

Complete step by step solution:
 We are given an equation $\cos p\theta =\cos q\theta ,p\ne q$ and we have to determine the general solution of this equation.
We will apply the theorem here and simplify to get the general solution of this equation. We will get,
$\begin{align}
  & p\theta =2n\pi \pm q\theta \\
 & p\theta \pm q\theta =2n\pi \\
 & \theta (p\pm q)=2n\pi \\
 & \theta =\dfrac{2n\pi }{p\pm q}
\end{align}$
The general solution of the trigonometric equation $\cos p\theta =\cos q\theta $ such that $p\ne q$,is $\theta =\dfrac{2n\pi }{p\pm q}$

Option ‘B’ is correct

Note: The domain of a function means the values of angles where the functions are defined while range means the resultant value of the function for that particular angle in the domain. In simple terms we can say that domain is input of the values and range is the output.
The domain of the trigonometric function cos is set of all of the real numbers while its range is $\left[ -1,1 \right]$. In a right angled triangle, it can be defined as the ratio of adjacent side to the hypotenuse of the triangle.
 The general solution of cos can be used to derive all the possible principal solutions for that question by substituting the value of $n$ so the solution we derived that is$\theta =\dfrac{2n\pi }{p\pm q}$will give all possible principal solutions for the equation $\cos p\theta =\cos q\theta $.
In this question direct theorem is applied so we have to learn and remember all the theorems to solve any of this kind of questions.