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Ex 3.4, 3 - cot x = - root 3, find principal and general - Ex 3.4


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Ex 3.4, 3 Find the principal and general solutions of the equation cot x = –√3 Given cot x = −√3 tan x = 1/cot⁡𝑥 tan x = 1/(−√3) tan x = (−1)/√3 We know that tan 30° = 1/√3 We find value of x where tan is negative tan is negative in llnd and lVth Quadrent Value in llnd Quadrant = 180° – 30° = 150° Value in lVth Quadrant = 360° – 30° = 330° So, principal Solution are x = 150° and x = 330° x = 150 × 𝜋/180 and x = 330 × 𝜋/180 x = 5𝜋/6 and x = 11𝜋/6 To find general solution Let tan x = tan y tan x = (−1)/√3 Form (1) and (2) tan y = (−1)/√3 tan y = tan 5𝜋/6 y = (5 𝜋)/6 Since tan x = tan y General Solution is x = nπ + y where n ∈ Z Put x = 5𝜋/6 Hence, x = nπ + 5𝜋/6 where n ∈ Z

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He provides courses for Mathematics from Class 9 to 12. You can contact him here.
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