JEE Main 2019Trigonometric EquationsTrigonometric EquationsEasyQuestionIf 0≤x<2π0 \le x < 2\pi 0≤x<2π, then the number of real values of xxx, which satisfy the equation cosx+cos2x+cos3x+cos4x=0\,\cos x + \cos 2x + \cos 3x + \cos 4x = 0cosx+cos2x+cos3x+cos4x=0 is:OptionsA7B9C3D5Check AnswerHide SolutionSolutioncosx+cos2x+cos3x+cos4x=0\cos x + \cos 2x + \cos 3x + \cos 4x = 0cosx+cos2x+cos3x+cos4x=0 ⇒\Rightarrow⇒ (cosx+cos3x)(\cos x + \cos 3x)(cosx+cos3x) + (cos2x+cos4x)(\cos 2x + \cos 4x)(cos2x+cos4x) = 0 ⇒2cos2xcosx+2cos3xcosx=0 \Rightarrow 2\cos 2x\cos x + 2\cos 3x\cos x = 0⇒2cos2xcosx+2cos3xcosx=0 ⇒2cosx(2cos5x2cosx2)=0 \Rightarrow 2\cos x\left( {2\cos {{5x} \over 2}\cos {x \over 2}} \right) = 0⇒2cosx(2cos25xcos2x)=0 cosx=0,cos5x2=0,cosx2=0\cos x = 0,\cos {{5x} \over 2} = 0,\cos {x \over 2} = 0cosx=0,cos25x=0,cos2x=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}x=π,2π,23π,5π,53π,57π,59π