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Ex 3.4, 8 - Find general solution of sec^2 2x = 1 - tan 2x - Teachoo

Ex 3.4, 8 - Chapter 3 Class 11 Trigonometric Functions - Part 2
Ex 3.4, 8 - Chapter 3 Class 11 Trigonometric Functions - Part 3
Ex 3.4, 8 - Chapter 3 Class 11 Trigonometric Functions - Part 4
Ex 3.4, 8 - Chapter 3 Class 11 Trigonometric Functions - Part 5
Ex 3.4, 8 - Chapter 3 Class 11 Trigonometric Functions - Part 6

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Ex 3.4, 8 Find the general solution of the equation sec2 2x = 1 – tan 2x sec2 2x = 1 – tan 2x 1 + tan2 2x = 1 – tan2x tan2 2x + tan2x = 1 – 1 tan2 2x + tan2x = 0 tan 2x (tan2x + 1) = 0 Hence We know that sec2 x = 1 + tan2 x So, sec2 2x = 1 + tan2 2x tan 2x = 0 tan 2x + 1 = 0 tan 2x = –1 We find general solutions for both separately General solution for tan 2x = 0 Let tan x = tan y tan 2x = tan 2y Also, tan 2x = 0 From (1) and (2) tan 2y = 0 We find solutions for tan 2x = 0 & tan 2x = –1 separately General solution for tan 2x = 0 Let tan x = tan y tan 2x = tan 2y Also, tan 2x = 0 From (1) and (2) tan 2y = 0 tan 2y = tan 0 2y = 0 y = 0 tan 2y = tan 0 2y = 0 y = 0 General solution is given by 2x = nπ + 2y 2x = nπ + 0 2x = nπ x = 𝒏𝝅/𝟐 General solution for tan 2x = −1 Let tan x = tan y tan 2x = tan 2y Given tan 2x = –1 From (3) and (4) tan 2y = –1 tan 2y = tan 3/4 π 2y = 3/4 π Rough We know that tan 45° = 1 But we need –1 So, angle is in 2nd and 4th quadrant θ = 45° 180 – θ = 180 – 45° = 135° = 135 × 𝜋/180 = 3/4 π General solution is given by 2x = nπ + 2y where n ∈ Z Putting 2y = 3/4 π 2x = nπ + 3/4 π x = 1/2 × ("nπ +" 3/4 " π" ) = 𝑛𝜋/2 + 3/8 π Hence, General Solutions are For tan 2x = 0 , x = 𝒏𝝅/𝟐 OR For tan 2x = −1 , x = 𝒏𝝅/𝟐 + 𝟑/𝟖 π Where n ∈ Z

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.