# If  0 ≤ x < 2 π, then the number of real values of x, which satisfy the equationcos x + cos 2x + cos 3x + cos 4x = 0, is : Option 1) 3 Option 2) 5 Option 3) 7 Option 4) 9

As we learnt in in

General Solution of Trigonometric Ratios -

$\cos \Theta = \cos \alpha$

$\Theta = 2n\pi \pm \alpha , n\epsilon I$

- wherein

$\alpha$ is the given angle

$\cos x+\cos 2x+\cos 3x+\cos 4x=0$

$\Rightarrow (\cos x+\cos 4x)+(\cos 2x+\cos 3x)=0$

$\Rightarrow 2\cos \frac{4x+x}{2}\:\:\cos\frac{4x-x}{2}+2\cos\:\frac{2x+3x}{2}\:\:\cos\frac{3x-2x}{2}=0$

$\Rightarrow 2\cos \frac{5x}{2}\:\:\left [ \cos \frac{3x}{2}+ \cos \frac{x}{2} \right ]=0$

$\Rightarrow 4\cos \frac{5x}{2}\:\:\cos x\:\:\cos \frac{x}{2}=0$

Thus,

$\cos \frac{5x}{2}=0\:\:\:\Rightarrow x=\frac{\pi}{5}, \frac{3\pi}{5}, \frac{7\pi}{5}, \frac{9\pi}{5}$                        $\left \{ \because x\varepsilon[0, 2\pi] \right$

$\cos x=0\:\:\:\Rightarrow x=\frac{\pi}{2}, \frac{3\pi}{2}$

$\cos \frac{x}{2}=0\:\:\Rightarrow x=\pi$

Hence, there are a total number of 7 solutions.

Option 1)

### 3

This option is incorrect.

Option 2)

### 5

This option is incorrect.

Option 3)

### 7

This option is correct.

Option 4)

### 9

This option is incorrect.

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