Solve the following: cos x + 3 sin x = 2

Given trigonometric equation is cos x + 3 sin x = 2 it is of form a sin x + b cos x = c . On dividing by both sides with a 2 + b 2 i.e., 3 2 + 1 2 = 2 , we get 1 2 cos x + 3 2 sin x = 2 2 ⇒ cos π 3 cos x + sin π 3 cos x = 1 2 ⇒ cos x - π 3 = cos π 4 ∵ cos ( A - B ) = cos A cos B + sin A sin B ⇒ x - π 3 = 2 n π ± π 4 , n ∈ Z ∵ cos x = cos α ⇒ x = 2 n π ± α , n ∈ Z ⇒ x = 2 n π + π 4 + π 3 or x = 2 n π - π 4 + π 3 Hence, the general solution of the given trigonometric equation is x = 2 n π + 7 π 12 or x = 2 n π + π 12 , n ∈ Z .

Solve the following: sec θ + tan θ = 3

Given trigonometric equation: sec θ + tan θ = 3 ⇒ 1 cos θ + sin θ cos θ = 3 ⇒ 1 + sin θ cos θ = 3 ⇒ 1 + sin θ = 3 cos θ ⇒ 3 cos θ - sin θ = 1 It is of form a sin x + b cos x = c . On dividing both sides by a 2 + b 2 i.e., - 1 2 + ( 3 ) 2 i.e., 2 , we get: 3 2 cos θ - 1 2 sin θ = 1 2 ⇒ cos θ cos π 6 - sin θ sin π 6 = 1 2 ⇒ cos θ + π 6 = 1 2 ∵ cos A cos B - sin A sin B = cos A + B ⇒ cos θ + π 6 = cos π 3 ⇒ θ + π 6 = 2 n π ± π 3 , n ∈ Z ∵ cos x = cos y ⇒ x = 2 n π ± y , n ∈ Z ⇒ θ + π 6 = 2 n π + π 3 or θ + π 6 = 2 n π - π 3 , n ∈ Z ⇒ θ = 2 n π + π 6 or θ = 2 n π - π 2 , n ∈ Z But tan θ , sec θ are not defined for θ = 2 n π - π 2 , n ∈ Z . Therefore, the general solution of sec θ + tan θ = 3 is θ = 2 n π + π 6 .

E M B I B E

Bureau's Higher Secondary Elements of Mathematics Vol.1>Chapter 4 - Trigonometric Functions>Exercises- 4(c)

Odisha Board Solutions for Chapter: Trigonometric Functions, Exercise 3: Exercises- 4(c)

Author:Odisha Board

Odisha Board Mathematics Solutions for Exercise - Odisha Board Solutions for Chapter: Trigonometric Functions, Exercise 3: Exercises- 4(c)

Attempt the free practice questions on Chapter 4: Trigonometric Functions, Exercise 3: Exercises- 4(c) with hints and solutions to strengthen your understanding. Bureau's Higher Secondary Elements of Mathematics Vol.1 solutions are prepared by Experienced Embibe Experts.