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    Maths-
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    Question

    e number of solutions of 2 sin to the power of 2 end exponent invisible function application x plus sin to the power of 2 end exponent invisible function application 2 x equals 2 comma blank x element of open square brackets 0 comma blank 2 pi close square brackets is

    1. 4  
    2. 5  
    3. 7  
    4. 6  

    The correct answer is: 6

      open parentheses 1 minus cos invisible function application 2 x close parentheses plus open parentheses 1 minus cos to the power of 2 end exponent invisible function application 2 x close parentheses equals 2
      rightwards double arrow cos invisible function application 2 x open parentheses cos invisible function application 2 x plus 1 close parentheses equals 0
      rightwards double arrow cos invisible function application 2 x equals 0 or cos invisible function application 2 x equals negative 1
      rightwards double arrow 2 x equals open parentheses 2 n plus 1 close parentheses pi divided by 2 or 2 x equals open parentheses 2 n plus-or-minus 1 close parentheses pi comma blank n element of Z
      rightwards double arrow x equals open parentheses 2 n plus 1 close parentheses pi divided by 4 orblank x equals open parentheses 2 n plus-or-minus 1 close parentheses pi divided by 2 comma blank n element of Z
      Hence, the solutions are pi divided by 4 comma blank 3 pi divided by 4 comma blank 5 pi divided by 4 comma blank 7 pi divided by 4 comma blank pi divided by 2 comma blank 3 pi divided by 2

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