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Trigonometry Equations: Solving for x in Different Quadrants

This guide explores trigonometric equations, showcasing methods to solve for x within specified ranges, including both degrees and radians. It details how to adapt reference angles for each quadrant, emphasizing the transformations needed for accurate solutions. Problem sets illustrate key concepts through various equations involving cosine, sine, and tangent functions, addressing both standard solutions and special cases when no valid solutions exist. Mastering these techniques enhances comprehension of trigonometric relationships and their applications.

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Trigonometry Equations: Solving for x in Different Quadrants

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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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