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Mathematics, five units · Ages 17–18

Solving Trigonometric Equations Across One Period

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Solve each equation for x over the interval 0° <= x < 360°. Show the primary solution, the symmetry step, and the final list of angles.

  1. Solve: sin(x) = 0.5

  2. Solve: cos(x) = 0

  3. Solve: tan(x) = 1

  4. Solve: 2cos(x) + sqrt(3) = 0

  5. Solve: 2sin^2(x) - sin(x) = 0

  6. Solve: cos(2x) = 0.5 for 0° <= x < 360°

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Solving Trigonometric Equations Across One Period

Answer key

  1. Solve: sin(x) = 0.5

    Primary angle: arcsin(0.5) = 30°. Second angle: 180° - 30° = 150°. Solutions: x = 30°, 150°.

  2. Solve: cos(x) = 0

    Primary angle: arccos(0) = 90°. Second angle: 360° - 90° = 270°. Solutions: x = 90°, 270°.

  3. Solve: tan(x) = 1

    Primary angle: arctan(1) = 45°. Tangent has a period of 180°, so the second angle is 45° + 180° = 225°. Solutions: x = 45°, 225°.

  4. Solve: 2cos(x) + sqrt(3) = 0

    Isolate cosine: cos(x) = -sqrt(3)/2. Primary angle: arccos(-sqrt(3)/2) = 150°. Second angle: 360° - 150° = 210°. Solutions: x = 150°, 210°.

  5. Solve: 2sin^2(x) - sin(x) = 0

    Factor: sin(x)(2sin(x) - 1) = 0. Either sin(x) = 0, giving x = 0°, 180°, or sin(x) = 0.5, giving x = 30°, 150°. Solutions: x = 0°, 30°, 150°, 180°.

  6. Solve: cos(2x) = 0.5 for 0° <= x < 360°

    Let u = 2x over [0°, 720°). Base angles for u are 60° and 300°. Adding 360° gives 420° and 660°. Divide each by 2: x = 30°, 150°, 210°, 330°.