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JEE Main202118 Mar 2021Morning ShiftMathematicsTrigonometric EquationsActual

The solutions of the equation 1 + sin 2 x sin 2 x sin 2 x cos 2 x 1 + cos 2 x cos 2 x 4 sin 2 x 4 sin 2 x 1 + 4 sin 2 x = 0 , 0 < x < π , are

Options

  1. Aπ 12 , π 6
  2. Bπ 6 , 5 π 6
  3. C5 π 12 , 7 π 12
  4. D7 π 12 , 11 π 12

Correct answer

D. 7 π 12 , 11 π 12

Step-by-step solution

1 + sin 2 x sin 2 x sin 2 x cos 2 x 1 + cos 2 x cos 2 x 4 sin 2 x 4 sin 2 x 1 + 4 sin 2 x = 0 use R 1 → R 1 + R 2 + R 3 ⇒ 2 + 4 sin 2 x 1 1 1 cos 2 x 1 + cos 2 x cos 2 x 4 sin 2 x 4 sin 2 x 1 + 4 sin 2 x = 0 use C 1 → C 1 - C 3     &   C 2 → C 2 - C 3   ⇒ 2 + 4 sin 2 x 0 0 1 0 1 cos 2 x - 1 - 1 1 + 4 sin 2 x = 0 ⇒ 2 + 4 sin 2 x = 0 ⇒ sin 2 x = - 1 2 ⇒ 2 x = π + π 6 , 2 π - π 6 x = π 2 + π 12 , π - π 12

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