solve for x if 7cos2x+2=0 and xE[0;360]

Question

I have no idea where to start with this question, do I use the same method as when finding  the general solution? Please help...

Answer 1

Yes, you'll use a similar method as you would to find the general solution. Begin by simplifying the equation a bit:

 

7cos2x + 2 = 0

7cos2x = -2

cos2x = -2/7

 

Now figure out all the angles on the unit circle where cosine is equal to -2/7. There should be two (one in Quadrant II and one in Quadrant III), since your cosine is negative. Taking the inverse cosine of -2/7 gives you one angle:

 

cos^-1(-2/7) = 106.6º

 

The other angle can be found by reflecting 106.6º over the x-axis to find its mirror image in Quadrant III. 106.6º is 73.4º less than 180º on the circle, so its reflection is 73.4º more than 180º, ie., 253.4º.

 

This means that 2x = 106.6º and 253.4º, so x = 53.3º and 126.7º. These are both solutions, as they both fall between 0º and 360º.

 

Since you haven't been asked to find a general solution, you should simply check the next few periodic values of 2x. At some point, you'll find that they begin to yield x-values that fall outside the 0º to 360º range:

 

2x = 106.6º + 360º = 466.6º, so x = 233.3º

2x = 466.6º + 360º = 826.6º, so x = 413.3º, but this is too large

 

2x = 253.4º + 360º = 613.4º, so x = 306.7º

2x = 613.4º + 360º = 973.4º, so x = 486.7º, but this is too large

 

So the set of values for x that satisfying the equation and fall within the given range is: {53.3º, 126.7º, 233.3º, 306.7º}.