If $$0 \le x < 2\pi $$, then the number of real values of $$x$$, which satisfy the equation $$\,\cos x + \cos 2x + \cos 3x + \cos 4x = 0$$ is:
JEE · Math · previous-year question
- A.7correct
- B.9
- C.3
- D.5
Answer
A. 7
Explanation
$$\cos x + \cos 2x + \cos 3x + \cos 4x = 0$$ $$ \Rightarrow $$ $$(\cos x + \cos 3x)$$ + $$(\cos 2x + \cos 4x)$$ = 0 $$ \Rightarrow 2\cos 2x\cos x + 2\cos 3x\cos x = 0$$ $$ \Rightarrow 2\cos x\left( {2\cos {{5x} \over 2}\cos {x \over 2}} \right) = 0$$ $$\cos x = 0,\cos {{5x} \over 2} = 0,\cos {x \over 2} = 0$$ $$x = \pi ,{\pi \over 2},{{3\pi } \over 2},{\pi \over 5},{{3\pi } \over 5},{{7\pi } \over 5},{{9\pi } \over 5}$$
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