JEE Main & Advanced Mathematics Trigonometric Equations Question Bank Solution of trigonometrical equations

  • question_answer
    \[\sin 6\theta +\sin 4\theta +\sin 2\theta =0,\]then \[\theta =\] [MP PET 1999; Pb. CET 2000]

    A) \[\frac{n\pi }{4}\]or \[n\pi \pm \frac{\pi }{3}\]

    B) \[\frac{n\pi }{4}\]or \[n\pi \pm \frac{\pi }{6}\]

    C) \[\frac{n\pi }{4}\]or \[2n\pi \pm \frac{\pi }{6}\]

    D) None of these

    Correct Answer: A

    Solution :

    \[\sin 6\theta +\sin 4\theta +\sin 2\theta =0\]           \[\Rightarrow 2\sin 4\theta \cos 2\theta +\sin 4\theta =0\] \[\Rightarrow \]\[\sin 4\theta (2\cos 2\theta +1)=0\] \[\Rightarrow \] \[2\cos 2\theta =-1\] \[\Rightarrow \] \[\cos 2\theta =-\frac{1}{2}\] \[\Rightarrow \] \[2\theta =2n\pi \pm \frac{2\pi }{3}\Rightarrow \theta =n\pi \pm \frac{\pi }{3}\] and \[\sin 4\theta =0\Rightarrow 4\theta =n\pi \Rightarrow \theta =\frac{n\pi }{4}\] \[\theta =\frac{n\pi }{4}\] or \[n\pi \pm \frac{\pi }{3}\].


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