J & K CET Engineering J and K - CET Engineering Solved Paper-2010

  • question_answer
    The general solution of the equation\[sin\text{ }x+cos\text{ }x=2\]is

    A)  \[x=n\pi \,+{{(-1)}^{n}}\,\frac{\pi }{4},\,n=....,-3,\,-2,\,\] \[-1,\,0,1,\,2,\,3,...\]

    B)  \[x=n\pi +{{(-1)}^{n}}\frac{\pi }{6},n=....,-3,-2,-1,\] \[0,\,1,\,2,\,3,....\]

    C)  \[x=n\pi +{{(-1)}^{n+1}}\frac{\pi }{4},n=...,-3,-2,\] \[-1,0,1,2,3,...\]

    D)  does not exist

    Correct Answer: D

    Solution :

    Given, equation is \[\sin x+\cos \,x=2\] \[\Rightarrow \] \[\cos x=2\,-\sin x\] \[\Rightarrow \] \[\sqrt{1-{{\sin }^{2}}x}=2-\sin x\] \[\Rightarrow \] \[1-{{\sin }^{2}}x=4+{{\sin }^{2}}x-4\,\sin x\] \[\Rightarrow \] \[2{{\sin }^{2}}x-4\,\sin x+3=0\] Here, \[{{B}^{2}}-4AC=16-24<0\] Therefore, solution of the equation does not exist because the equation have complex roots.


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