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

  • question_answer
    If \[2{{\cos }^{2}}x+3\sin x-3=0,\,\,0\le x\le {{180}^{o}}\], then x = [MP PET 1986]

    A) \[{{30}^{o}},{{90}^{o}},{{150}^{o}}\]

    B) \[{{60}^{o}},{{120}^{o}},{{180}^{o}}\]

    C) \[{{0}^{o}},{{30}^{o}},{{150}^{o}}\]

    D) \[{{45}^{o}},{{90}^{o}},{{135}^{o}}\]

    Correct Answer: A

    Solution :

    \[2-2{{\sin }^{2}}x+3\sin x-3=0\] \[\Rightarrow \]\[(2\sin x-1)\,(\sin x-1)=0\]\[\Rightarrow \] \[\sin x=\frac{1}{2}\] or\[\sin x=1\] \[\Rightarrow \] \[x=\frac{\pi }{6},\,\frac{5\pi }{6}\], \[\frac{\pi }{2}\] i.e., \[30{}^\circ ,\text{ 150}{}^\circ \text{, 90}{}^\circ \].


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