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JEE Main & Advanced Mathematics Trigonometric Identities Question Bank Mock Test - Trigonometric Functions

  • question_answer
    The number of solutions of \[\sin x+\sin 2x+\sin 3x=\cos x+\cos 2x+\cos 3x,\]\[0\le x\le 2\pi \], is

    A) 7                     

    B) 5

    C) 4                     

    D) 6

    Correct Answer: D

    Solution :

    [d] we have \[(sinx+sin3x)+sin2x=(cosx+cos3x)+cos2x\] Or \[2\sin 2x\cos x+\sin 2x=2\cos 2x\cos x+\cos 2x\] Or \[\sin 2x(2cosx+1)=cos2x(2cosx+1)\] Or \[(2cosx+1)(sin2x-\cos 2x)=0\] Or \[\cos x=-1/2\sin 2x-\cos 2x=0\] \[\Rightarrow x=2n\pi \pm (2\pi /3)or\,tan2x=1=tan(\pi /4)\] \[=2n\pi \pm (2\pi /3)or\,x=(4n+1)\pi /8,n\in Z\] But here \[0\le x\le 2\pi .\]Hence, \[x=\pi /8,\,\,5\pi /8,\,\,2\pi /3,\,\,9\pi /8,\,\,4\pi /3,\,\,13\pi /8.\]


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