2 cos(theta-13)= square root of 3

Question

The question asks to solve algebraically. The restrictions are 0 through 720 degrees.

Answer 1

Let t represent theta. 2cos(t-13)=sqrt(3); cos(t-13)=sqrt(3)/2. cos(30)=sqrt(3)/2 so t-13=30 and t=43 degrees. But cos(-30)=cos(30)=cos(360-30)=cos(330). Therefore, t-13=330 and t=343 degrees. Also cos(30)=cos(360+30)=cos(390); cos(330)=cos(330+360)=cos(690). So t-13=390 and 690, making t=403 and 703. There are four solutions: theta=43, 343, 403 and 703.