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

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°

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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