1. Class 11
2. Important Question for exams Class 11

Transcript

Ex 3.4, 4 Find the principal and general solution of cosec x = –2 Given cosec x = –2 ⇒ 1/sin⁡𝑥 = -2 sin x = (−1)/2 We know that sin 30° = 1/2 But we need negative value of sin x sin x is negative in lllrd & lVth Quadrant Value in IIIrd Quadrant = 180° + 30° = 210° Value in IVth Quadrant = 360° – 30° = 330° So, Principle Solution are x = 210° and x = 330° x = 210 × 𝜋/180 and x = 330 × 𝜋/180 x = 7𝜋/6 and x = 11𝜋/6 To find general solution Let sin x = sin y sin x = – 1/2 From (1) and (2) sin y = – 1/2 sin y = sin 7𝜋/6 y = 7𝜋/6 Since sin x = sin y General Solution is x = nπ + ( -1)n y where n ∈ Z Put y = 7𝜋/6 Hence, x = nπ +(-1)n 7𝜋/6 Where n ∈ Z

Class 11
Important Question for exams Class 11
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