Question

What values for theta (0 less than or equal to theta less than equal to 2pi) satisfy the equation?

Answer 1

Solve this as you would an equation that does not involve trig.  Don't let the trig scare you.  If you had to solve 2x+8=0, the first thing you would do is factor out the common 2.  In our equation, we have a common cos theta.  I'm going to use beta as my angle.  When we factor out beta, here's what we have. $cos \beta (2sin \beta +1)=0$.  The Zero Product Property tells us that at least one of those factors has to equal zero.  So we set them both equal to zero and solve.  Let's get the equations first, then we will need our unit circle.  First equation set to equal zero is $cos \beta =0$.  On our unit circle, cos is the value inside the parenthesis that is in the x position within our coordinate.  Look at all those coordinates as you go around the unit circle once (once around is equivalent to 2pi).  You will find that the the cos is 0 at $\frac{ \pi }{2}, \frac{3 \pi }{2}$.  The next equation is $2sin\beta +1=0$.  Move the 1 over by subtraction and divide by 2 to get $sin \beta =- \frac{1}{2}$.  Same as before, go around the unit circle one time and look to see where the coordinate in the y place is -1/2.  Sin corresponds to the y coordinate.  You will find that sin is -1/2 at $\frac{7 \pi }{6}, \frac{11 \pi }{6}$.  And there you go!  Trig is so much fun!!!