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Trigonometry—Equations (cont’d)

SIN. COS. /2. 1.57. 0. 0. –1. 1. . 0, 2 . 180°. 0°, 360°. 3.14. 0, 6.28. –1. 0. 3 /2. 270°. 4.71. x = 180 – 56.3 = 123.7  = 2.16. x = 360 – 56.3 = 303.7  = 5.30. Trigonometry—Equations (cont’d). Page 38. In Degrees. In Radians.

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Trigonometry—Equations (cont’d)

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  1. SIN COS /2 1.57 0 0 –1 1  0, 2 180° 0°, 360° 3.14 0, 6.28 –1 0 3/2 270° 4.71 x = 180 – 56.3 = 123.7  = 2.16 x = 360 – 56.3 = 303.7  = 5.30 Trigonometry—Equations (cont’d) Page38 In Degrees In Radians In terms of  In terms of decimal 90° • If we know ref and  is in the respective quadrants we can find  as follows: Q I:  = refQ II:  = 180 – refQ III:  = 180 + refQ IV:  = 360 – ref • Problems: • Solve for x where 0° x < 360°: • 1. cos 2x = ¼ 2. cos 2x – cos x – 2 = 0 • Solve for x where 0 x < 2: • cos 2x – 3sin x + 4 = 0 2. 2tan2 x + tan x – 3 = 0 3. 2sec2x + 3sec x + 1 = 0 • 1 – 2 sin2 x – 3sin x + 4 = 0 (2tan x + 3)(tan x – 1) = 0 • –2sin2 x – 3sin x + 5 = 0 tan x = –3/2 | tan x = 1 • 2sin2 x + 3sin x – 5 = 0 xref = tan–1(3/2) | xref = tan–1(1) • (2sin x + 5)(sin x – 1) = 0 xref = 56.3 | xref = 45 • sin x = –5/2 | sin x = 1x  Q II and QIV | x  Q I and QIII •  x = 90 • = x = 45 = (No solutions because sin x can’t be < –1 nor > 1) x = 180 + 45 = 225 =

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