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General
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Question

If 0 less than theta less than 2 pi, then the intervals of values of theta for which 2 sin squared space theta minus 5 sin space theta plus 2 greater than 0, is

  1. open parentheses 0 comma pi over 6 close parentheses union open parentheses fraction numerator 5 pi over denominator 6 end fraction comma 2 pi close parentheses
  2. open parentheses pi over 8 comma fraction numerator 5 pi over denominator 6 end fraction close parentheses
  3. open parentheses 0 comma pi over 6 close parentheses union open parentheses pi over 6 comma fraction numerator 5 pi over denominator 6 end fraction close parentheses
  4. open parentheses fraction numerator 41 pi over denominator 48 end fraction comma pi close parentheses

The correct answer is: open parentheses 0 comma pi over 6 close parentheses union open parentheses fraction numerator 5 pi over denominator 6 end fraction comma 2 pi close parentheses

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To solve a trigonometric inequation of the type sin x ≥ a where |a| ≤ 1, we take a hill of length 2straight pi  in the sine curve and write the solution within that hill. For the general solution, we add 2nstraight pi. For instance, to solve sin space x greater or equal than negative 1 half, we take the hill open square brackets negative pi over 2 comma fraction numerator 3 pi over denominator 2 end fraction close square brackets over which solution is negative pi over 6 less than X less than fraction numerator 7 pi over denominator 6 end fraction The general solution is 2 straight n pi minus pi over 6 less than straight x less than 2 straight n pi plus fraction numerator 7 pi over denominator 6 end fraction, n is any integer. Again to solve an inequation of the type sin x ≤ a, where |a| ≤ 1, we take a hollow of length 2straight pi in the sine curve. (since on a hill, sinx ≤ a is satisfied over two intervals). Similarly cos x ≥ a or cosx ≤a, |a| ≤ 1 are solved.

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In the interval left square bracket negative pi divided by 4 comma pi divided by 2 right square bracket, the equation, cos space 4 x plus fraction numerator 10 tan space x over denominator 1 plus tan squared space x end fraction equals 3 has

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