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The number of solutions of the equation sin to the power of 3 end exponent invisible function application x cos invisible function application x plus sin to the power of 2 end exponent invisible function application x cos to the power of 2 end exponent invisible function application x plus sin invisible function application x cos to the power of 3 end exponent invisible function application x equals 1 comma blankin the interval open square brackets 0 comma blank 2 pi close square brackets comma blankis

  1. 4  
  2. 2  
  3. 1  
  4. 0  

The correct answer is: 0


    Then given equation can be written as
    sin invisible function application x cos invisible function application x open square brackets sin to the power of 2 end exponent invisible function application x plus sin invisible function application x cos invisible function application x plus cos to the power of 2 end exponent invisible function application x close square brackets equals 1
    rightwards double arrow sin invisible function application x cos invisible function application x open square brackets 1 plus sin invisible function application x cos invisible function application x close square brackets equals 1
    rightwards double arrow sin invisible function application 2 x open square brackets 2 plus sin invisible function application 2 x close square brackets equals 4
    rightwards double arrow sin invisible function application 2 x equals fraction numerator negative 2 plus-or-minus square root of 4 plus 16 end root over denominator 2 end fraction equals negative 1 plus-or-minus square root of 5
    Which is not possible

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