Home About Authors Look Inside The Book Answers Ordering Feedback Contact Us
 
 

 
 

Answers To The Exercises: 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14
Lesson 11
 


11.1.
Equations of types cos= A and sin= A are called simplest.

11.2.
The trig functions are periodic ones. Therefore, for infinitely many arguments, they have the same value.

11.3.
sin = 2.

11.4

11.5.
1) Factor the equation: sin x(sin x – 1) = 0. It can be broken into two simplest equations sin x = 0 and sin x = 1. The solution can be described by the two sets:
x = n and x = /2 + 2n.

2) Factor the equation: (2cos x – 1)(cos x + 1) = 0. It can be broken into two equations: cos x = ½ and cos x = –1. The solution is described by two sets:
x = ± /3 + 2n and x = + 2n.

3) Factor the equation: (sin x + 2)(sin x – 3) = 0. From here we get two equations:
sin x = –2 and sin x = 3. None of them has a solution. Therefore,
the original equation does not have solutions.

4) Substitute sin2 x = 1 – cos2 x. The equation becomes cos2 x – 2cos x – 3 = 0 or
(cos x + 1)(cos x – 3) = 0. Two simplest equations are cos x = –1 and cos x = 3.
The second equation does not have solutions. The first equation gives the solutions
x = + 2n.

5) The equation can be broken into two equations sin x = 1 and sin x = –1. The solutions are x = /2 + n.

6) The equation can be broken into two equations tan x = 1 and tan x = –1. The solutions are x = ± /4 +n. These solutions can also be written in the form:
x = /4 + n/2.

7) Factor: sin x(2sin2 x – 1) = 0. From here, sin x = 0 and sin2 x = ½. The first equation gives x = n. The second equation gives x = ± (–1)n /6 + n.

8) Factor: (2sin2 x – 1)( sin2 x – 1) = 0. From here, sin2 x = ½ and sin2 x = 1. The first equation gives x = /4 + n/2. The second equation gives x = /2 + n.

9) Substitute cot x = cos x / sin x. The equation can be reduced to 2sin2 x – 3cos x = 0. Substitute sin2 x = 1 – cos2 x. The equation becomes 2cos2 x + 3cos x – 2 = 0 or (2cos x – 1)(cos x +2) = 0. From here, cos x = ½ and cos x = – 2. The second equation does not have solutions. The first equation gives x = ± /3 + 2n.

10) Use the formula sin 2x = 2sin x·cos x. The equation can be written as
cos x(2sin x – 1) = 0. from here, cos x = 0 and sin x = ½. The first equation gives
x = /2 + n. The second equation gives x = (–1)n/6 + n.

11.6.
The equation 2sin(3t + 1) = 0 has the solutions t=(1/3)(n - 1). The first three positive solutions are t = (– 1)/3, t = (2– 1)/3 and t = (3– 1)/3.

 

 

  Copyright © 2003 Dad's Lessons. All Rights Reserved