auto generate barcode vb net 3 cos2 x in .NET

Draw QR in .NET 3 cos2 x

14.5 3 cos2 x
Denso QR Bar Code Scanner In Visual Studio .NET
Using Barcode Control SDK for .NET Control to generate, create, read, scan barcode image in .NET framework applications.
QR Code Encoder In .NET Framework
Using Barcode encoder for VS .NET Control to generate, create Denso QR Bar Code image in .NET applications.
sin2 x.
Denso QR Bar Code Reader In .NET
Using Barcode scanner for .NET Control to read, scan read, scan image in .NET applications.
Generate Barcode In Visual Studio .NET
Using Barcode creation for .NET Control to generate, create barcode image in VS .NET applications.
cos2 x, we have 3 cos2 x /3, 2 /3, 4 /3, and 5 /3. 1 cos2 x or 4 cos2 x 1. Then cos x
Decode Barcode In .NET
Using Barcode recognizer for .NET Control to read, scan read, scan image in .NET framework applications.
Paint QR-Code In C#
Using Barcode encoder for Visual Studio .NET Control to generate, create QR Code 2d barcode image in VS .NET applications.
First Solution. Replacing sin2 x by 1 1/2 and the required solutions are x
QR-Code Generation In .NET Framework
Using Barcode creator for ASP.NET Control to generate, create QR Code JIS X 0510 image in ASP.NET applications.
QR-Code Printer In VB.NET
Using Barcode printer for .NET Control to generate, create Denso QR Bar Code image in Visual Studio .NET applications.
Second Solution. Dividing the equation by cos2 x, we have 3 above are obtained.
Print UCC-128 In VS .NET
Using Barcode drawer for .NET Control to generate, create UCC-128 image in .NET framework applications.
Matrix Barcode Generator In .NET
Using Barcode generator for Visual Studio .NET Control to generate, create 2D Barcode image in .NET applications.
tan2 x. Then tan x
Drawing Linear In .NET
Using Barcode maker for .NET framework Control to generate, create 1D image in .NET applications.
Code11 Maker In .NET Framework
Using Barcode printer for VS .NET Control to generate, create Code 11 image in VS .NET applications.
23, and the solutions
EAN13 Scanner In Visual Studio .NET
Using Barcode reader for .NET framework Control to read, scan read, scan image in .NET framework applications.
1D Barcode Creator In VB.NET
Using Barcode printer for Visual Studio .NET Control to generate, create 1D Barcode image in .NET applications.
14.6 2 sin x
Code 128B Creation In Java
Using Barcode generation for Android Control to generate, create Code 128 image in Android applications.
Code 39 Full ASCII Encoder In Visual Basic .NET
Using Barcode maker for Visual Studio .NET Control to generate, create ANSI/AIM Code 39 image in Visual Studio .NET applications.
csc x
Draw Barcode In None
Using Barcode creator for Office Word Control to generate, create bar code image in Office Word applications.
GTIN - 128 Encoder In None
Using Barcode generation for Online Control to generate, create GTIN - 128 image in Online applications.
1 sin x, and rearranging, we have 1)(sin x 1) 0 1, x 1 /2. 2 sin x
Printing ANSI/AIM Code 39 In Java
Using Barcode maker for BIRT Control to generate, create Code 39 Extended image in Eclipse BIRT applications.
GS1-128 Generation In Java
Using Barcode maker for BIRT reports Control to generate, create UCC - 12 image in Eclipse BIRT applications.
Multiplying the equation by sin x to get 2 sin2 x sin x 1
(2 sin x
From 2 sin x Check: For x For x
0, sin x
1/2 and x
7 /6 and 11 /6; from sin x 1 2( 1/2) ( 2)
/2, 2 sin x csc x 2(1) 1 7 /6 and 11 /6, 2 sin x csc x /2, 7 /6, and 11 /6.
The solutions are x
14.7 2 sec x
tan x
cot x.
2 cos x sin x cos x cos x sin x
Transforming to sines and cosines and clearing of fractions, we have or 2 sin x sin 2x cos 2x 1
Then sin x
1/2 and x
/6 and 5 /6.
14.8 tan x
3 cot x
Multiplying by tan x and rearranging, we have tan2 x 4 tan x 3 (tan x 1)(tan x 3) 0. From tan x 1 0, tan x 1 and x /4 and 5 /4; from tan x 3 0, tan x 3 and x 1.25 and 4.39. Check: For x For x /4 and 5 /4, 1.25 and 4.39, tan x tan x 3 cot x 3 cot x 1 3(1) 4 3.0096 3(0.3323) 4.0065 < 4
The solutions are /4, 1.25, 5 /4, and 4.39.
CHAPTER 14 Trigonometric Equations
14.9 csc x
cot x
23 cot x and squaring, we have cot 2 x 2 0. Then cot x 1/ 23 and x /3
First Solution. Writing the equation in the form csc x csc 2 x Replacing csc2 x by 1 and 4 /3. Check: For x For x 3 2 23 cot x
cot2 x and combining, we get 2 23 cot x p/3, csc x 4 /3, csc x /3. cot x cot x 2/ 23 2/ 23 1/ 23 23
1> 23 2 23
The required solution is x
Second Solution. Upon making the indicated replacement, the equation becomes 1 sin x cos x sin x 23 and, clearing of fractions, cos2 x 2 cos x 3 sin2 x 2 3(1 1 cos x 23 sin x
Squaring both members, we have 1
2 cos x 4 cos2 x
cos2 x) or 1)(cos x 1) 0
2(2 cos x
From 2 cos x 1 0, cos x 1/2 and x /3 and 5 /3; from cos x 1 0, cos x 1 and x . Now x /3 is the solution. The values x and 5 /3 are to be excluded since csc is not defined while csc 5 /3 and cot 5 /3 are both negative.
14.10 cos x
23 sin x
1 1 23 sin x and squaring, we have 3 sin2 x 3(1 cos2 x) cos x
Putting the equation in the form cos x 2 cos x
then, combining and factoring, 4 cos2 x From 2 cos x Check: 1 For x For x For x 0, cos x 0, cos x 2 cos x 1/2 and x 23 sin x 23 sin x 23 sin x 0 and 4 /3. 2 2(2 cos x 23(0) 1/2 1/2 1)(cos x 1) 1 0 0, cos x 1 and x 0.
2 /3 and 4 /3; from cos x 1 1
2p/3, cos x 4p/3, cos x
23 A 23/2 B 2 1 23 A 23/2 B 1
The required solutions are x
14.11 2 cos x
sin x.
4 cos2 x 4(1 5 sin x
As in Prob. 14.10, we obtain 1 1 2 sin x 2 sin x sin2 x sin2 x 1) 0 3.78 and 5.64; from sin2 x) 3
2 sin x 3/5 /2. 1 1 1
(5 sin x
3)(sin x
From 5 sin x 3 0, sin x sin x 1 0, sin x 1 and x Check: For x For x For x /2, 2(0)
0.6000, the reference angle is 0.64, and x
3.78, 2( 0.8021) 5.64, 2(0.8021) < 1
( 0.5972) ( 0.5972)
The required solutions are x
/2 and 5.64.
CHAPTER 14 Trigonometric Equations
Equations Involving Multiple Angles and Half Angles 14.12 sin 3x
Then and Each of these values is a solution.
1 2 22.
Since we require x such that 0 3x x
x < 2 , 3x must be such that 0
3x < 6 .
5 /4, 7 /4, 13 /4, 15 /4, 21 /4, 23 /4 5 /12, 7 /12, 13 /12, 5 /4, 7 /4, 23 /12
14.13 cos 2x
Then
1 2x
1 2.
Since we require x such that 0 p/3 and x 2 /3.
x < 2 , 2x must be such that 0
1 2x
14.14 sin 2x
cos x
Substituting for sin 2x, we have 2 sin x cos x cos x cos x (2 sin x 1) 0. From cos x 0, x /2 and 3 /2; from sin x 1/2, x 7 /6 and 11 /6. The required solutions are x /2, 7 /6, 3 /2, and 11 /6.
1 14.15 2 cos2 2 x
cos2 x.
1 First Solution. Substituting 1 cos x for 2 cos 22x, the equation becomes cos2 x cos x 1 0; then 1 25 0.6180 and obtain 1.6180, 0.6180. Since cos x cannot exceed 1, we consider cos x cos x 2 the solutions x 2.24 and 4.04. 1 Second Solution. To solve 22 cos 1 x cos x and 22 cos 2 x cos x, we square and obtain the equation 2 of this problem. The solution of the first of these equations is 4.04 and the solution of the second is 2.24.
Copyright © OnBarcode.com . All rights reserved.