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JEE Main 2019
Trigonometric Equations
Trigonometric Equations
Easy

Question

If 0x<2π0 \le x < 2\pi , then the number of real values of xx, which satisfy the equation cosx+cos2x+cos3x+cos4x=0\,\cos x + \cos 2x + \cos 3x + \cos 4x = 0 is:

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Solution

cosx+cos2x+cos3x+cos4x=0\cos x + \cos 2x + \cos 3x + \cos 4x = 0 \Rightarrow (cosx+cos3x)(\cos x + \cos 3x) + (cos2x+cos4x)(\cos 2x + \cos 4x) = 0 2cos2xcosx+2cos3xcosx=0 \Rightarrow 2\cos 2x\cos x + 2\cos 3x\cos x = 0 2cosx(2cos5x2cosx2)=0 \Rightarrow 2\cos x\left( {2\cos {{5x} \over 2}\cos {x \over 2}} \right) = 0 cosx=0,cos5x2=0,cosx2=0\cos x = 0,\cos {{5x} \over 2} = 0,\cos {x \over 2} = 0 x=π,π2,3π2,π5,3π5,7π5,9π5x = \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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