Question

If cos x, cos 2x, cos 3x are in A.P, then (where \[n\in 1\]) –
(a) \[x=\left( 2n+1 \right)\dfrac{\pi }{4}\]
(b) \[x=n\pi \]
(c) \[x=\left( 2n+1 \right)\pi \]
(d) \[x=2n\pi \]

Answer 1

Hint: We know that when three terms, say a, b and c are in A.P then 2b = a + c, substitute the value of a, b and c as cos x, cos 2x and cos 3x respectively. Now use \[\cos A+\cos B=2\cos \left( \dfrac{A+B}{2} \right).\cos \left( \dfrac{A-B}{2} \right)\].
Now, simplify the equation to find the value of x by using the trigonometric equations.

Complete step-by-step answer:
 If cos x, cos 2x and cos 3x are in A.P, then we have to find the value of x. We know that if 3 terms say a, b and c are in A.P, then we get, 2b = a + c.
Therefore, if cos x, cos 2x and cos 3x are in A.P, then we get,
\[2\left( \cos 2x \right)=\cos 3x+\cos x\]
We know that,
\[\cos A+\cos B=2\cos \left( \dfrac{A+B}{2} \right).\cos \left( \dfrac{A-B}{2} \right)\]
By using this in the RHS of the above equation, we get,
\[2.\left( \cos 2x \right)=2\cos \left( \dfrac{3x+x}{2} \right)\cos \left( \dfrac{3x-x}{2} \right)\]
\[\Rightarrow 2\left( \cos 2x \right)=2\cos \left( 2x \right)\cos \left( x \right)\]
By transposing all the terms to the LHS, we get,
\[2\left( \cos 2x \right)-2\left( \cos 2x \right).\left( \cos x \right)=0\]
By taking out 2cos 2x common from the above equation, we get,
\[2\cos 2x\left( 1-\cos x \right)=0\]
Hence, we get,
\[2\cos 2x=0\]
\[\cos 2x=0\]
\[\Rightarrow 1-\cos x=0\]
\[\cos x=1\]
Let us take cos 2x = 0. We know that \[\cos \dfrac{\pi }{2}=0\], so we get,
\[\cos 2x=\cos \dfrac{\pi }{2}\]
We know that, when \[\cos \theta =\cos \alpha \], then,
\[\theta =2n\pi \pm \alpha \]
By using this in the above equation, we get,
\[2x=2n\pi \pm \dfrac{\pi }{2}\]
\[\Rightarrow x=n\pi \pm \dfrac{\pi }{4}\]
\[\Rightarrow x=\left( 4n+1 \right)\left( \dfrac{\pi }{4} \right)\text{ and }x=\left( 4n-1 \right)\left( \dfrac{\pi }{4} \right)\]
Let us take cos x – 1. We know that when \[\cos \theta =\cos \alpha \], then \[\theta =2n\pi \pm \alpha \] and we know that cos 0 = 1. So, we get,
\[\cos x=\cos 0\]
\[\Rightarrow x=2n\pi \pm 0=2n\pi \]
Therefore, we get \[x=\left( 2n+1 \right)\dfrac{\pi }{4}\] or \[x=2n\pi \].
Hence, option (a) and (d) are the right answers.

Note: Here students must remember that if \[\cos \theta =1\], then \[\theta =2n\pi \] , and if \[\cos \theta =0\], then \[\theta =\left( 2n+1 \right)\dfrac{\pi }{2}\] because these results are very useful while solving the trigonometric equations. Also, take care that in the last, in one solution, the angle is 2x and in the other it is x. So, it is advisable to take care of this and don’t write it incorrectly.