Solve for x where-360 s x s 360. sincx-22)- 1 3. tan(x+ 40")-1
Trigonometry (11th Edition)
11th Edition
ISBN:9780134217437
Author:Margaret L. Lial, John Hornsby, David I. Schneider, Callie Daniels
Publisher:Margaret L. Lial, John Hornsby, David I. Schneider, Callie Daniels
Chapter1: Trigonometric Functions
Section: Chapter Questions
Problem 1RE:
1. Give the measures of the complement and the supplement of an angle measuring 35°.
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![Solve for x where-360 s xs 360.
and sin 1 110° = sin(30° + 3 x 360) sin 3
You can summarise the solutions to trigor
13
Vodacom SA 6l
DO 0I 67/ 19:22
B/s
Abuti Amo
Today, 12:19
EXERCISE 16
1.
sin(x-22)-
3.
tan(x+ 40") -1
cos x+ COs 71° = sin 17
6.
7.
V2 sin(x- 27") -1 0
8.
9.
Scos(x+ 15)-2=0
11
3sin(x + 32) = 2sin 32
for
The concept of a general so
You need not restrict the solutions to trigono
10°; 360°) or [-360°; 360°]
Look at the solutions to sin x =
sin 30° =, but so does sin 390° = sin(30° +
sin 750 sin(30° + 2 x 360) =
Sin 150°- but so does sin 510° = sIn(1o
sin 870
sin(150 + 2 x 360°) = 3 and sin
O
>](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F5de3b3f4-5a2a-49a9-ab03-f84aa1c39cee%2F0f8db48d-49b1-4bfc-afdf-33919330b296%2Fzaji7i_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Solve for x where-360 s xs 360.
and sin 1 110° = sin(30° + 3 x 360) sin 3
You can summarise the solutions to trigor
13
Vodacom SA 6l
DO 0I 67/ 19:22
B/s
Abuti Amo
Today, 12:19
EXERCISE 16
1.
sin(x-22)-
3.
tan(x+ 40") -1
cos x+ COs 71° = sin 17
6.
7.
V2 sin(x- 27") -1 0
8.
9.
Scos(x+ 15)-2=0
11
3sin(x + 32) = 2sin 32
for
The concept of a general so
You need not restrict the solutions to trigono
10°; 360°) or [-360°; 360°]
Look at the solutions to sin x =
sin 30° =, but so does sin 390° = sin(30° +
sin 750 sin(30° + 2 x 360) =
Sin 150°- but so does sin 510° = sIn(1o
sin 870
sin(150 + 2 x 360°) = 3 and sin
O
>
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