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📐 Mathematics  ·  Trigonometric Functions  ·  JEE

General solution of 2cos²x + 3sinx = 0 is:

Answer: x = 7π/6 + 2nπ or x = 11π/6 + 2nπ.

  • A x = nπ + (-1)<sup>n</sup> × 7π/6, an incorrect general form
  • B x = nπ + (-1)<sup>n</sup> × (-π/6), using the wrong reference angle
  • C x = 7π/6 + 2nπ or x = 11π/6 + 2nπ
  • D No solution exists within the given range of values

Correct answer: C. x = 7π/6 + 2nπ or x = 11π/6 + 2nπ

Explanation: 2(1-sin²x) + 3sinx = 0. 2sin²x - 3sinx - 2 = 0. (2sinx+1)(sinx-2)=0. sinx = -1/2 (sinx=2 impossible). x = 7π/6 + 2nπ or x = 11π/6 + 2nπ.

xy0 deg30 deg45 deg (1/sqrt2, 1/sqrt2)60 deg90 degcos45sin45

Unit circle with standard angles 0, 30, 45, 60, 90 degrees; at 45 degrees the point is (cos45, sin45) = (1/sqrt2, 1/sqrt2).

Concept context

Ratios, identities, and applications in triangles

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