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JEE Main & Advanced Mathematics Trigonometric Identities Question Bank Self Evaluation Test - Trigonometric Function

  • question_answer
    The least difference between the roots, in the first quadrant \[\left( 0\le x\le \frac{\pi }{2} \right),\] of the equation \[4\cos x(2-3{{\sin }^{2}}x)+(\cos 2x+1)=0\]is

    A) \[\frac{\pi }{6}\]            

    B) \[\frac{\pi }{4}\]   

    C) \[\frac{\pi }{3}\]            

    D) \[\frac{\pi }{2}\]

    Correct Answer: A

    Solution :

    We have, \[4\cos x(2-3{{\sin }^{2}}x)+(\cos 2x+1)=0\] \[\Rightarrow \,\,4\cos x(3{{\cos }^{2}}x-1)+2{{\cos }^{2}}x=0\] \[\Rightarrow \,\,2\cos x(6{{\cos }^{2}}x+\cos x-2)=0\] \[\Rightarrow \,\,2\cos x(3\cos x+2)(2cosx-1)=0\] \[\Rightarrow \]either \[\cos \,\,x=0\] which gives \[x=\pi /2\] or \[\cos \,x=-2/3\] Which gives no value of x for which \[0\le x\le \pi /2\] or  \[cos\,x=1/2,\]which gives \[x=\pi /3\] So, the required difference \[=\pi /2-\pi /3=\pi /6\]


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