Trigonometric Equations in Visual Studio .NET

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CHAPTER 14 Trigonometric Equations
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EXAMPLE 14.6 Solve 4 sin x
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3. 4 sin x 3 sin x 3/4 0.75 0.85 3.14 0.85 2.29.
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The reference angle is 0.85 and the solutions for x are x (See Chap. 6 to review the use of reference angles.) Check: For x For x 0.85, 4 sin 0.85 2.29, 4 sin 2.29
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0.85 and x
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4(0.7513) 3.0052 < 3 4[sin (3.14 2.29)] 4[sin 0.85]
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4[0.7513]
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3.0052 < 3
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If a calculator is used, sin 2.29 is computed directly, so 4 sin 2.29 4(0.7523) Thus, the solutions to the nearest hundredth radian are 0.85 and 2.29.
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(NOTE: Since the checks used approximate numbers, the symbol < was used to indicate the result was approximately equal to the value needed.)
EXAMPLE 14.7 Solve 15 cos2 x
7 cos x
2/3 0.6667 and cos x 1/5 0.2. 15 cos x 7 cos x 2 0, (3 cos x 2)(5 cos x 1) 0, and cos x From cos x 0.6667, the reference angle is 0.84 and x 0.84 3.14 0.84 2.3 and x 0.84 3.14 0.84 3.98. From cos x 0.2, the reference angle is 1.37 and x 1.37 and x 2 1.37 6.28 1.37 4.91. Thus, to the nearest hundredth radian the solutions for x are 1.37, 2.3, 3.98, and 4.91. (E) Equation contains a multiple angle.
EXAMPLE 14.8 Solve cos 2x
3 sin x
0. 2 sin2 x 3 sin x 2 0, 2 sin2 x 3 sin x 2 0,
cos 2x 3 sin x (2 sin x 1)(sin x
1 2)
0, (1 2 sin x) 3 sin x 1 0, 1 0, and sin x 2 and sin x 2. 1 sin x
From sin x 2 , x /6 and 5 /6. From sin x 2, there are no solutions since The solutions for x are /6 and 5 /6.
EXAMPLE 14.9 Solve 2 cos2 2x
1 for all x.
cos 2x.
1 2.
2 cos2 2x cos 2x, 2 cos2 2x cos 2x 0, cos 2x (2 cos 2x 1) 0, and cos 2x 0 and cos 2x Since we want 0 x < 2 , we find all values of 2x such that 0 2x < 4 . From cos 2x 0, 2x /2, 3 /2, 5 /2, and 7 /2 and x /4, 3 /4, 5 /4, and 7 /4. 1 From cos 2x 2, 2x /3, 5 /3, 7 /3, and 11 /3 and x /6, 5 /6, 7 /6, and 11 /6. Thus the required angles x are /6, /4, 3 /4, 5 /6, 7 /6, 5 /4, 7 /4, and 11 /6. (F) Equations containing half angles.
EXAMPLE 14.10 Solve 4 sin2 2 x
1. 2(1 cos x)/2]
First Solution. 4
sin2 1 x 2
1, 4[
1, 2
2 cos x
1, cos x
1 2,
and x
/3 and 5 /3.
The required solutions are x Second Solution. 4 sin2 1 x 2 for 1x such that 0 2 From sin 1x 2 From sin
1 2x 1 2x 1 1 2, 2x 1 2,
/3 and 5 /3.
1 1, sin2 2 x 1 4,
and sin 1x 2 /3 and 5 /3.
1 2.
Since we want 0
x < 2 , we want all solutions
p. p/6 and 5 /6, and x /3 and 5 /3. there are no solutions since sin 1x 2 0 for all x such that 0
1 2x
The required solutions are x
CHAPTER 14 Trigonometric Equations
SOLVED PROBLEMS
Solve each of the trigonometric equations in Probs. 14.1 to 14.19 for all x such that 0 x < 2 . (If all solutions are required, add 2n , where n is zero or any positive or negative integer, to each result given.) In a number of the solutions, the details of the check have been omitted. 14.1 2 sin x 1 0.
1/2 and x /6 and 5 /6.
Here sin x
14.2 sin x cos x
From sin x 0, x 0 and ; from cos x 0, x /2 and 3 /2. The required solutions are x 0, /2, , and 3 /2.
14.3 (tan x
1)(4 sin2 x
0, sin x 23/2 and x /3,
From tan x 1 0, tan x 1 and x /4 and 5 /4; from 4 sin2 x 3 2 /3, 4 /3, and 5 /3. The required solutions are x /4, /3, 2 /3, 5 /4, 4 /3, and 5 /3.
14.4 sin2 x
sin x
1 0, sin x 1 and x /2.
Factoring, (sin x 2)(sin x 1) 0. From sin x 2 0, sin x 2 and there is no solution; from sin x The required solution is x /2.
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