Sender: Pierre Boivin trigonometric equation, extending solution. 2cos 5cos3 = 0 for between 360 and 720 degree When I factor[ 2cos (square)  5cos 3], I get (2cos + 1)(cos  3). 2cos + 1 = 0, 2cos = 1, cos = 0.5,. Using inv cos on calculator, I get 120 degree related angle. When I graph I get two values, between 90 and 180 degree and between 180 and 270 degrees. How do I find those two values. How do use 120 degree in relation with the x axis. Maybe my factoting is wrong.
thank you in advance Hi Pierre, Your factoring is fine. The difficulty arises in trying to solve cos x = 1/2. I would quickly sketch a graph of one cycle of y = cos x, for x from 0^{o} to 360^{o}.
Notice that the graph crosses the xaxis at 90^{o} and 270^{o}. If you draw horizontal lines ay y = 1/2 and y = 1/2 you see that there are two values of x, at Q and R, where cos x = 1/2 and two values of x, at P and S, where cos x = 1/2.
The key to finding Q and R is to first find P. Either by using the inverse cosine on your calculator or by using the "306090" triangle you can find that P = 60^{o}. Now from the symetry of the diagram P is 30^{o} to the left of the intersection point at 90^{o} so Q is 30^{o} to the right of 90^{o}. Thus Q is 90^{o} + 30^{o} = 120^{o}. For the other pair of points R is 30^{o} to the left of 270^{o} and S is 30^{o} to the right. Thus R is 270^{o}  30^{o} = 240^{o} and S is 270^{o} + 30^{o} = 300^{o}. Hence cos x = 1/2 at 120^{o} and at 240^{o}.
Cheers,
