Question 7 - Principal and General Solutions - Chapter 3 Class 11 Trigonometric Functions
Last updated at April 16, 2024 by Teachoo
Principal and General Solutions
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Principal and General Solutions
Last updated at April 16, 2024 by Teachoo
Question 7 Find the general solution of the equation sin 2x + cos x = 0 sin 2x + cos x = 0 Putting sin 2x = 2 sin x cos x 2 sin x cos x + cos x = 0 cos x (2sin x + 1) = 0 Hence, We find general solution of both equations separately cos x = 0 2sin x + 1 = 0 2sin x = â1 sin x = (â1)/2 General solution for cos x = 0 Given cos x = 0 General Solution is x = (2n + 1) đ /đ where n â Z General solution for sin x = (âđ)/đ Let sin x = sin y Given sin x = (â1)/2 From (1) and (2) sin y = (â1)/2 sin y = sin 7đ/6 y = 7đ/6 Rough We know that sin 30Â° = 1/2 But we need (â1)/2 So, angle is in 3rd & 4th quadrant Î¸ = 30Â° 180 + Î¸ = 180 + 30 = 210Â° = 210 Ã đ/180 = 7/6 Ī General Solution is x = nĪ + (â1)n y where n â Z Putting y = 7đ/6 x = nĪ + (â1)n 7đ/6 Where n â Z Therefore, For cos x = 0, x = (2n + 1) đ /đ OR For sin x = (â1)/2, x = nĪ + (â1)n đđ /đ where n â Z