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Unformatted text preview: MAC1114  Homework 10
Due Tuesday  March 22 • For each of the following equations:
(a) Find all solutions.
(b) Solve in 1. 3 tan3 x = tan x.
(a) 3 tan3 x−tan x = 0 ⇒ tan x(3 tan2 x−1) = 0 ⇒ (b) Plug in
2. x = 0, 1 to get tan x = 0
⇒
3 tan2 x − 1 = 0 tan x = 0
√
tan x = ± 33 x = nπ, n ∈ Z ⇒ x = π + nπ, n ∈ Z .
6 π
x = 56 + nπ, n ∈ Z π
π
{0, π , 56 π, 76 , 11π }.
6
6 sec2 4x − 1 = 3. (a) 4x 4x
sec2 4x = 4 ⇒ sec 4x = ±2 ⇒ cos 4x = ± 1 ⇒
2
4x 4x (b) Plug in
3. [0, 2π ). x = 0, 1, 2, 3 to get π
π
{ 12 , 5π , π , π , 7π , 11π , 23 ,
12 6 3 12 12 π
π
x = 12 + nπ
3 + 2nπ
2 x = 5π + nπ
5π
+ 2nπ
3
12
2
⇒
2π
x = π + nπ 3 + 2nπ
6
2 4π
x = π + nπ
3 + 2nπ
3
2
5π 13π 17π 7π 4π 19π 23π 5π 11π
, 12 , 12 , 6 , 3 , 12 , 12 , 3 , 6 }.
6 =
=
=
= 2 sin x + csc x = 0
1
1
2 sin x = − csc x ⇒ 2 sin x = − sin x ⇒ sin2 x = − 2 . This has no solution since the lefthand side is always positive.
4. 2 sin2 x − 7 sin x + 3 = 0
(a) Set u = sin x (ignore part (b) for this one) to get 2u2 −7u+3 = 0 ⇒ (2u−1)(u−3) = 0 ⇒ 2 sin x − 1 = 0
⇒
sin x = 3 1 sin x = 1
2 ⇒ x = π + 2nπ, x = 11π + 2mπ.
6
6
sin x = 3 ...
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This note was uploaded on 06/10/2011 for the course MAC 1114 taught by Professor Gentimis during the Spring '11 term at University of Florida.
 Spring '11
 Gentimis
 Algebra, Equations

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