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Over left square bracket negative pi comma pi right square bracket the solution of 2 sin squared invisible function application open parentheses x plus pi over 4 close parentheses plus square root of 3 cos invisible function application 2 x greater or equal than 0 is

  1. open square brackets fraction numerator negative 5 pi over denominator 6 end fraction comma fraction numerator 5 pi over denominator 6 end fraction close square brackets
  2. left square bracket 0 comma pi right square bracket
  3. open square brackets negative pi comma fraction numerator negative 7 pi over denominator 12 end fraction close square brackets union open square brackets negative pi over 4 comma fraction numerator 5 pi over denominator 12 end fraction close square brackets union open square brackets fraction numerator 3 pi over denominator 4 end fraction comma pi close square brackets
  4. left square bracket negative pi comma pi right square bracket

The correct answer is: open square brackets negative pi comma fraction numerator negative 7 pi over denominator 12 end fraction close square brackets union open square brackets negative pi over 4 comma fraction numerator 5 pi over denominator 12 end fraction close square brackets union open square brackets fraction numerator 3 pi over denominator 4 end fraction comma pi close square brackets




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    Let f left parenthesis x right parenthesis equals fraction numerator c o s e c to the power of 4 end exponent invisible function application x minus 2 c o s e c to the power of 2 end exponent invisible function application x plus 1 over denominator c o s e c invisible function application x left parenthesis c o s e c invisible function application x minus s i n invisible function application x right parenthesis plus fraction numerator s i n invisible function application x minus c o s invisible function application x over denominator sin invisible function application x end fraction plus c o t invisible function application x end fraction The sum of all the solutions of f (x) = 0 in [0, 100p] is

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