Solve the following equations: sin x + cos x = 2

Given s i n x + c o s x = 2 ⇒ 1 2 s i n x + 1 2 c o s x = 1 ⇒ s i n π 4 s i n x + c o s π 4 c o s x = 1 ∵ s i n π 4 = c o s π 4 = 1 2 ⇒ c o s ( x - π 4 ) = 1 ⇒ cos ( x - π 4 ) = cos 0 The general solution of cos x = cos a is x = 2 n π ± a , n ∈ Z . ⇒ x - π 4 = 2 n π ± 0 , n ∈ Z ⇒ x = 2 n π + π 4 ; n ∈ Z

Solve the following equations: 2 sin 2 x + 2 cos 2 x = 2 2

Given 2 s i n 2 x + 2 c o s 2 x = 2 2 ⇒ 2 s i n 2 x + 2 1 - s i n 2 x = 2 2 ⇒ 2 s i n 2 x + 2 2 s i n 2 x = 2 2 Let 2 s i n 2 x = y ∴ y + 2 y = 2 2 ⇒ y 2 - 2 2 y + 2 = 0 ⇒ ( y - 2 ) 2 = 0 ⇒ y = 2 ⇒ 2 s i n 2 x = 2 1 2 (replace y ) ⇒ s i n 2 x = 1 2 (Since bases are equal, powers are also equal) ⇒ s i n 2 x = 1 2 2 = s i n 2 π 4 The general solution of sin 2 x = sin 2 a is x = n π ± a , n ∈ Z . ⇒ x = n π ± π 4 ; n ∈ Z

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MATHEMATICS for CLASS XI>Chapter 11 - Trigonometric Equations>EXERCISE 11.1

R D Sharma Solutions for Chapter: Trigonometric Equations, Exercise 1: EXERCISE 11.1

Author:R D Sharma

R D Sharma Mathematics Solutions for Exercise - R D Sharma Solutions for Chapter: Trigonometric Equations, Exercise 1: EXERCISE 11.1

Attempt the practice questions on Chapter 11: Trigonometric Equations, Exercise 1: EXERCISE 11.1 with hints and solutions to strengthen your understanding. MATHEMATICS for CLASS XI solutions are prepared by Experienced Embibe Experts.