Download presentation
Presentation is loading. Please wait.
Published byBrook Gordon Modified over 9 years ago
1
Solving Trig Equations Starter CStarter C Starter C SolutionsStarter C Solutions Starter DStarter D Starter D SolutionsStarter D Solutions
2
Starter A 1) Sin(x) = 0.82 2) Cos(x) = 0.49 3) Tan(x) = 0.14 4) Sin(x) = 0.38 Solve these equations. 0° ≤ x ≤ 360° Home ‘Achieve’
3
Starter A Soln 1) Sin(x) = 0.82 2) Cos(x) = 0.49 3) Tan(x) = 0.14 4) Sin(x) = 0.38 Solve these equations. 0° ≤ x ≤ 360° Home 180 0, 360 A TC S 180 0, 360 A TC S 180 0, 360 A TC S 180 0, 360 A TC S
4
Starter B 2) Tan(x) = -2.4 0° ≤ x ≤ 360° Solve these equations Home ‘Achieve’ 3) Sin(x) = 0.77 0 ≤ x ≤ 2π 1) Cos(x) = -0.34 0° ≤ x ≤ 360° 4) Cos(x) = 0.14 0 ≤ x ≤ 2π
5
Starter B Solutions 2) Tan(x) = -2.4 0° ≤ x ≤ 360° Home 3) Sin(x) = 0.77 0 ≤ x ≤ 2π 1) Cos(x) = -0.34 0° ≤ x ≤ 360° 4) Cos(x) = 0.14 0 ≤ x ≤ 2π 180 0, 360 A TC S 180 0, 360 A TC S π 0, 2 π A TC S π A TC S
6
Starter C 1) 4Sin(x) = 3.2 0° ≤ x ≤ 360° 2) Cos(x) + 3 = 3.19 0 ≤ x ≤ 2π 3) 0.5Tan(x) = 0.24 0 ≤ x ≤ 2π 4) 3Sin(x) + 1 = -1.8 0° ≤ x ≤ 360° Solve these equations. Home
7
Starter C Solutions 1) 4Sin(x) = 3.2 0° ≤ x ≤ 360° 2) Cos(x) + 3 = 3.19 0 ≤ x ≤ 2π 3) 0.5Tan(x) = 0.24 0 ≤ x ≤ 2π 4) 3Sin(x) + 1 = -1.8 0° ≤ x ≤ 360° Home 180 0, 360 A TC S π 0, 2 π A TC S π A TC S 180 0, 360 A TC S
8
Starter D Sketch these graphs. Home 1) y = 2Sin(x) + 1 0° ≤ x ≤ 360° 2) y = Cos(x) + 3 0 ≤ x ≤ 2π 3) y = 0.5Tan(x) 0 ≤ x ≤ 2π 4) y = 3Sin(x) + 1 0° ≤ x ≤ 360°
9
Starter E 1) Cos x = 0.42 2) 4Cos x = 3.2 3) Tan x + 1 = 2.5 4) Sin(x + 20) = 0.6 Solve these equations. 0° ≤ x ≤ 360° Home ‘Achieve’ 5) Sin 2x = 0.3 6) 4Cos x + 14 = 15 (I st solution only) 7) Sin (x – 1) = -0.85 8) 2Tan x + 8 = 9.4 (I st solution only) 0 ≤ x ≤ 2π For # 5 - 8 ‘Merit’
10
Starter E Solutions 1) Cos x = 0.42 2) 4Cos x = 3.2 3) Tan x + 1 = 2.5 4) Sin(x + 20) = 0.6 Solve these equations. 0° ≤ x ≤ 360° Home ‘Achieve’ 5) Sin 2x = 0.3 6) 4Cos x + 14 = 15 (I st solution only) 7) Sin (x + 1) = -0.85 8) 2Tan x + 8 = 9.4 (I st solution only) 0 ≤ x ≤ 2π For # 5 - 8 x = 65.2° or 294.8° x = 36.9° or 323.1° x = 56.3° or 236.3° x = 16.9° or 106.9° x = 0.15 or 1.42 or 3.29 or 4.56 x = 1.32 or 4.97 4Cos x = 1 x = 3.16 or 4.27 x = 2.90 or 6.04 180 0, 360 A TC S Positive Cos so quadrant 1 & 4 360 – 65.2° = 294.8° ‘Merit’ Cos x = 0.8 Tan x = 1.5 Let ‘A’ = (x + 20) Let ‘A’ = 2x Cos x = 0.25
Similar presentations
© 2025 SlidePlayer.com. Inc.
All rights reserved.