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

  • question_answer
    a If \[4{{\sin }^{2}}\theta +2(\sqrt{3}+1)\cos \theta =4+\sqrt{3}\], then the general value  of \[\theta \]is

    A) \[2n\pi \pm \frac{\pi }{3}\]

    B) \[2n\pi +\frac{\pi }{4}\]

    C) \[n\pi \pm \frac{\pi }{3}\]

    D) \[n\pi -\frac{\pi }{3}\]

    Correct Answer: A

    Solution :

    \[4-4{{\cos }^{2}}\theta +2\,(\sqrt{3}+1)\cos \theta =4+\sqrt{3}\] \[\Rightarrow \] \[4{{\cos }^{2}}\theta -2\,(\sqrt{3}+1)\cos \theta +\sqrt{3}=0\] \[\Rightarrow \] \[\cos \theta =\frac{2(\sqrt{3}+1)\pm \sqrt{4{{(\sqrt{3}+1)}^{2}}-16\sqrt{3}}}{8}\] \[\Rightarrow \] \[\cos \theta =\frac{\sqrt{3}}{2}\text{ or}\,\,\text{1/2}\Rightarrow \theta =2n\pi \pm \frac{\pi }{6}\] or \[2n\pi \pm \pi /3\].


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