Question

The trigonometric equation $ \cos 2x-3\cos x+1=\dfrac{1}{\left( \cot 2x-\cot x \right)\sin \left( x-\pi \right)} $ holds if \[\]
A. $ \cos x=0 $ \[\]
B. $ \cos x=1 $ \[\]
C. $ \cos x=\dfrac{5}{2} $ \[\]
D. None of these \[\]

Answer 1

Hint: We multiply both side by $ \sin \left( x-\pi \right) $ and use shift by $ \pi $ formula $ \sin \left( \pi -\theta \right)=-\sin \theta $ . We convert all the rest of the terms into cosines using double angle formula $ \cos 2\theta =2{{\cos }^{2}}-1,\tan 2\theta =\dfrac{2\tan \theta }{1-{{\tan }^{2}}\theta } $ and the Pythagorean trigonometric identity $ {{\sec }^{2}}\theta -{{\tan }^{2}}\theta =1 $ . We get an equation in $ \cos x $ which we solve for $ \cos x $ to choose the correct option. \[\]

Complete step-by-step answer:
We are given the following trigonometric equation in the question.
 \[\cos 2x-3\cos x+1=\dfrac{1}{\left( \cot 2x-\cot x \right)\sin \left( x-\pi \right)}\]
We are asked in the question for what condition on $ \cos x $ the above equation holds true. We know from domain and range of trigonometric functions that the cosine function $ \cos x $ lies between $ -1 $ to 1. Let us multiply $ \sin \left( x-\pi \right) $ both side to have;

 \[\Rightarrow \left( \sin \left( x-\pi \right) \right)\left( \cos 2x-3\cos x+1 \right)=\dfrac{1}{\cot 2x-\cot x}\]
We use the formula for shift by $ {{\pi }^{c}} $ of sine that is $ \sin \left( \theta -\pi \right)=-\sin \theta $ in the left hand side to have;
 \[\Rightarrow -\sin x\left( \cos 2x-3\cos x+1 \right)=\dfrac{1}{\cot 2x-\cot x}\]
We use the cosine double angle formula $ \cos 2\theta =2{{\cos }^{2}}\theta -1 $ in the left hand side of above step for $ \theta =x $ to have;
 \[\begin{align}
  & \Rightarrow -\sin x\left( 2{{\cos }^{2}}x-1-3\cos x+1 \right)=\dfrac{1}{\cot 2x-\cot x} \\
 & \Rightarrow -\sin x\left( 2{{\cos }^{2}}x-3\cos x \right)=\dfrac{1}{\cot 2x-\cot x} \\
\end{align}\]
We convert the co-tangents into tangents in the right hand side using the reciprocal relation $ \cot \theta =\dfrac{1}{\tan \theta } $ to have;
  \[\Rightarrow -\sin x\left( 2{{\cos }^{2}}x-3\cos x \right)=\dfrac{1}{\dfrac{1}{\tan 2x}-\dfrac{1}{\tan x}}\]
We use tangent double angle formula $ \tan 2\theta =\dfrac{2\tan \theta }{1-{{\tan }^{2}}\theta } $ for $ \theta =x $ in the right hand side of the above step to have;
 \[\begin{align}
  & \Rightarrow -\sin x\left( 2{{\cos }^{2}}x-3\cos x \right)=\dfrac{1}{\dfrac{1}{\dfrac{2\tan x}{1-{{\tan }^{2}}x}}-\dfrac{1}{\tan x}} \\
 & \Rightarrow -\sin x\left( 2{{\cos }^{2}}x-3\cos x \right)=\dfrac{1}{\dfrac{1-{{\tan }^{2}}x}{2\tan x}-\dfrac{1}{\tan x}} \\
\end{align}\]
We multiply 2 in the numerator and denominator $ \dfrac{1}{\tan x} $ in the right hand side to have;
 \[\begin{align}
  & \Rightarrow -\sin x\left( 2{{\cos }^{2}}x-3\cos x \right)=\dfrac{1}{\dfrac{1-{{\tan }^{2}}x}{2\tan x}-\dfrac{2}{2\tan x}} \\
 & \Rightarrow -\sin x\left( 2{{\cos }^{2}}x-3\cos x \right)=\dfrac{1}{\dfrac{1-{{\tan }^{2}}x-2}{2\tan x}} \\
 & \Rightarrow -\sin x\left( 2{{\cos }^{2}}x-3\cos x \right)=\dfrac{1}{\dfrac{-1-{{\tan }^{2}}x}{2\tan x}} \\
 & \Rightarrow -\sin x\left( 2{{\cos }^{2}}x-3\cos x \right)=\dfrac{2\tan x}{-\left( 1+{{\tan }^{2}}x \right)} \\
\end{align}\]
We use the Pythagorean identity of tangent and secant of angle $ {{\sec }^{2}}\theta -{{\tan }^{2}}\theta =1 $ in the above step to have;
  \[\Rightarrow -\sin x\left( 2{{\cos }^{2}}x-3\cos x \right)=\dfrac{2\tan x}{-{{\sec }^{2}}x}\]
We convert the tangent and secant in the right hand side using the identities $ \sec \theta =\dfrac{1}{\cos \theta },\tan \theta =\dfrac{\sin \theta }{\cos \theta } $ to have;
 \[\Rightarrow -\sin x\left( 2{{\cos }^{2}}x-3\cos x \right)=\dfrac{2\dfrac{\sin x}{\cos x}}{-\dfrac{1}{{{\cos }^{2}}x}}\]
We cancel out the like terms top have;
 \[\begin{align}
  & \Rightarrow 2{{\cos }^{2}}x-3\cos x=2\cos x \\
 & \Rightarrow 2{{\cos }^{2}}x-5\cos x=0 \\
\end{align}\]
Let us factorize by taking $ \cos x $ common and have;
 \[\begin{align}
  & \Rightarrow \cos x\left( 2\cos x-5 \right)=0 \\
 & \Rightarrow \cos x=0\text{ or }2\cos x-5=0 \\
 & \Rightarrow \cos x=0\text{ or }\cos x=\dfrac{5}{2} \\
\end{align}\]
We have two possibilities $ \cos x=0 $ or $ \cos x=\dfrac{5}{2}=2.5 $ but we know that $ -1\le \cos x\le 1 $ for all $ x\in \mathsf{\mathbb{R}} $ . So we reject $ \cos x=\dfrac{5}{2} $ and accept $ \cos x=0 $ as the only possible condition for the given equation to hold true. So the only correct option is A. \[\]

So, the correct answer is “Option A”.

Note: We can further solve for $ x $ from the obtained condition $ \cos x=0 $ and find solutions as $ x=\left( 2n+1 \right)\dfrac{\pi }{2} $ where $ n $ is an arbitrary integer. We note that the question presumes $ \cot 2x-\cot x\ne 0 $ and $ \sin \left( x-\pi \right)\ne 0\Rightarrow \sin x\ne 0 $ and that is why we could cancel out $ -\sin x $ . We must be careful of the confusion between reduction formula $ \sin \left( \pi -\theta \right)=\sin \theta $ and shift formula $ \sin \left( \theta -\pi \right)=-\sin \theta $ .