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If 4 cos2 x° - 1 = 0 and 0 < x < 90, find(i) x(ii) sin2 xº + cos2 x® (iii) cos2 x° - sin2 xº. |
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Answer» cos²(2x) = 3/4 cos(2x) = ± √(3/4) = ±√3/2 Remember, cos(Θ) = ±√3/2 when Θ = 30°, 150° So, 2x = 30° and 2x = 150° x = 15° and x = 75° both meet the CRITERION 0 ≤ x ≤ 90 2. 2 sin²(x) - sin(x) = 0 sin(x) * (2 sin(x) - 1) = 0 So EITHER sin(x) = 0 or 2SIN(x) - 1 = 0 sin(x) = 0 when x = 0° or 180° 2sin(x) - 1 = 0 --> sin(x) =1/2 sin(x) = 1/2 when x = 30° or x = 150° To satisfy the criterion 0 ≤ x ≤ 90, x = 0° or 30° Hope it helps Follow me Mark as brainlist |
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