# SOLUTION: solve. Give all positive values of the angle between 0degrees and 360degrees. Give any approximate value to the nearest minute only. a.) cos 4x= sin2x and if you can th

Algebra ->  Algebra  -> Trigonometry-basics -> SOLUTION: solve. Give all positive values of the angle between 0degrees and 360degrees. Give any approximate value to the nearest minute only. a.) cos 4x= sin2x and if you can th      Log On

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 Question 117922: solve. Give all positive values of the angle between 0degrees and 360degrees. Give any approximate value to the nearest minute only. a.) cos 4x= sin2x and if you can this is a problem I also had trouble with b.) 3 sin(theta)-4 cos (theta)= 2 Answer by Edwin McCravy(8909)   (Show Source): You can put this solution on YOUR website!```solve. Give all positive values of the angle between 0° and 360°. Give any approximate value to the nearest minute only. a.) cos4x = sin2x Use the identity: cos(2A) = 1 - 2sin²A and the fact that 4x = 2(2x) to replace the left side of cos 4x = sin2x cos[2(2x)] = sin2x 1 - 2sin²2x = sin2x -2sin²2x - sin2x + 1 = 0 2sin²2x + sin2x - 1 = 0 (2sin2x - 1)(sin2x + 1) = 0 2sin2x - 1 = 0 sin2x + 1 = 0 2sin2x = 1 sin2x = -1 sin2x = 1/2 To get all values for x between 0° and 360°, we must take 2x between 0° and 720°: 2x = 30°, 150°, 390°, 510° 2x = 270°, 630° x = 15°, 75°, 195°, 255° x = 135°, 315° --------------------------------------------------- and if you can this is a problem I also had trouble with b.) 3 sinq - 4 cosq = 2 The best way to handle a linear combination of sine and cosine, such as m·sinq ± n·cosq is 1. To draw a right triangle with legs m and n, 2. Find an acute angle of that triangle. 3. Solve for m and n in terms of that acute angles of the triangle: In this case m=3 and n=4, so draw the triangle: We calculate agle F using tan(F) = or F = 53.13010235° We find the hypotenuse to be 5 by the Pythagorean theorem: c² = 3² + 4² c² = 9 + 16 c² = 25 c = 5 Now 3 = m = 5cosF, and 4 = n = 5sinF so substitute for the 3 and the 4 3·sinq - 4·cosq = 2 5cosF·sinq - 5sinF·cosq = 2 Now divide through by 5 cosF·sinq - sinF·cosq = Now use the identity sin(A-B) = sinA·cosB - cosA·sinB, reversing the factors in the two terms on the left side: sinq·cosF - cosq·sinq = sin(q-F) = .4 Now since 0° < q < 360° 0°-F < q-F < 360°-F and since F = 53.13010235° 0 - 53.13010235° < q-F < 360° - 53.13010235° -53.13010235° < q-F < 306.8608976° So q-F = 23.57617646°, 156.4218215° q = 23.57617646°+F, 156.4218215°+F q = 23.57617646°+53.13010235°, 156.4218215°+53.13010235° q = 76.70627881°, 209.5519239° To change to minutes, multiply the decimal parts by 60 (.70627881)(60) = 42.3767286 minutes or 42 minutes to the nearest minute (.5519239)(60) = 33.115434 minutes or 33 minutes to the nearest minute So the answers are: q = 76°42', 209°33' Edwin```