Question

If \[0\le x\le 2\pi \], then the real values of x, which satisfy the equation \[\cos x+\cos 2x+\cos 3x+\cos 4x=0\].
(a) 3
(b) 5
(c) 7
(d) 9

Answer 1

Hint: Group \[\cos x\] with \[\cos 3x\] and \[\cos 2x\] with \[\cos 4x\] and use the formula: - \[\cos A+\cos B=2\cos \left( \dfrac{A+B}{2} \right)\cos \left( \dfrac{A-B}{2} \right)\] to simplify the expression. Take the common terms together and apply the same formula to write all the cosine functions as the product. Substitute each term equal to 0 and use the formula: - If \[\cos a=0,a=\left( 2n+1 \right)\dfrac{\pi }{2},n\in I\], to write the general solution. Substitute the value of ‘n’ so that x lies between 0 and \[2\pi \] and count the total values of x obtained to get the answer.

Complete step by step answer:
Here, we have been provided with the equation: -
\[\Rightarrow \cos x+\cos 2x+\cos 3x+\cos 4x=0\]
Grouping \[\cos x\] with \[\cos 3x\] and \[\cos 2x\] with \[\cos 4x\], we get,
\[\Rightarrow \left( \cos x+\cos 3x \right)+\left( \cos 4x+\cos 2x \right)=0\]
Applying the identity: - \[\cos A+\cos B=2\cos \left( \dfrac{A+B}{2} \right)\cos \left( \dfrac{A-B}{2} \right)\], we get,
\[\begin{align}
  & \Rightarrow 2\cos \left( \dfrac{3x+x}{2} \right)\cos \left( \dfrac{3x-x}{2} \right)+2\cos \left( \dfrac{4x+2x}{2} \right)\cos \left( \dfrac{4x-2x}{2} \right)=0 \\
 & \Rightarrow 2\cos 2x\cos x+2\cos 3x\cos x=0 \\
 & \Rightarrow 2\cos x\left( \cos 2x+\cos 3x \right)=0 \\
\end{align}\]
Again, applying the identity: - \[\cos A+\cos B=2\cos \left( \dfrac{A+B}{2} \right)\cos \left( \dfrac{A-B}{2} \right)\], we get,
\[\Rightarrow 2\cos x\times 2\cos \left( \dfrac{3x+2x}{2} \right)\cos \left( \dfrac{3x-2x}{2} \right)=0\]
\[\begin{align}
  & \Rightarrow 8\cos x\cos \dfrac{5x}{2}\cos \dfrac{x}{2}=0 \\
 & \Rightarrow \cos \dfrac{x}{2}\cos x\cos \dfrac{5x}{2}=0 \\
\end{align}\]
Substituting each term equal to 0, we get,
1. \[\cos \dfrac{x}{2}=0\]
We know that the general solution of the trigonometric equation: - \[\cos a=0\] is given as: - \[a=\left( 2n+1 \right)\dfrac{\pi }{2},n\in I\].
\[\begin{align}
  & \Rightarrow \cos \dfrac{x}{2}=0 \\
 & \Rightarrow \dfrac{x}{2}=\left( 2n+1 \right)\dfrac{\pi }{2} \\
 & \Rightarrow x=\left( 2n+1 \right)\pi \\
\end{align}\]
Now, we have been given that \[0\le x\le 2\pi \]. Therefore, we have,
For \[n=0\Rightarrow x=\pi \].
Hence, \[x=\pi \].
2. \[\cos x=0\]
\[\Rightarrow x=\left( 2n+1 \right)\dfrac{\pi }{2}\]
To satisfy the condition \[0\le x\le 2\pi \], we have,
For n = 0 \[\Rightarrow x=\dfrac{\pi }{2}\]
For n = 1 \[\Rightarrow x=\dfrac{3\pi }{2}\]
Hence, \[x=\dfrac{\pi }{2},\dfrac{3\pi }{2}\].
3. \[\cos \dfrac{5x}{2}=0\]
\[\begin{align}
  & \Rightarrow \dfrac{5x}{2}=\left( 2n+1 \right)\dfrac{\pi }{2} \\
 & \Rightarrow x=\left( 2n+1 \right)\dfrac{\pi }{5} \\
\end{align}\]
To satisfy the condition, \[0\le x\le 2\pi \], we have,
For n = 0 \[\Rightarrow x=\dfrac{\pi }{5}\]
For n = 1 \[\Rightarrow x=\dfrac{3\pi }{5}\]
For n = 2 \[\Rightarrow x=\dfrac{5\pi }{5}=\pi \]
For n = 3 \[\Rightarrow x=\dfrac{7\pi }{5}\]
For n = 4 \[\Rightarrow x=\dfrac{9\pi }{5}\]
Hence, \[x=\dfrac{\pi }{5},\dfrac{3\pi }{5},\pi ,\dfrac{7\pi }{5},\dfrac{9\pi }{5}\]
Clearly, we can see that \[x=\pi \] is common in case (1) and case (3), so we will count it only once. Therefore, on counting all the different values of x obtained, we get,
Total number of values of x = 7.

So, the correct answer is “Option c”.

Note: One may note that we have grouped \[\cos x\] with \[\cos 3x\] and \[\cos 2x\] with \[\cos 4x\]. There is not any reason for that, you can group any two terms together because at last you will get the same expression. Remember that we have to substitute each trigonometric function equal to 0 and find the value of x. Before counting all the values of x obtained, check if any value is coming more than one, if yes then count it only once.