    1. Chapter 3 Class 11 Trigonometric Functions
2. Serial order wise
3. Ex 3.4

Transcript

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 So, tan 2x = 0 or tan 2x + 1 = 0 tan 2x = 0 or tan 2x = – 1 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 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 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 ) x = 1/2 × (nπ + 3/4 π ) x = 𝑛𝜋/2 + 3/8 π where n ∈ Z Hence, General Solution are For tan 2x = 0 , x = 𝑛𝜋/2 For tan 2x = -1 , x = 𝑛𝜋/2 + 3/8 π Where n ∈ Z

Ex 3.4 