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  • Please,please help me 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 :

 If  0 ≤ x < 2 π, then the number of real values of x, which satisfy the equation

cos x + cos 2x + cos 3x + cos 4x = 0, is :

  • Option 1)

    3

  • Option 2)

    5

  • Option 3)

    7

  • Option 4)

    9

 

Answers (1)

best_answer

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.

Posted by

prateek

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