What values for theta (0 less than or equal to theta less than equal to 2pi) satisfy the equation? 2 sin theta cos theta + sqrt 3 cos theta = 0 the answer is pi/2, 4pi/3, 3pi/2, 5pi/3 i just need to know how to get it

for  cosine theta can be factored out. all it is doing is multiplying with multiple terms after all.    now we apply the zero product property.the zero product property allows us to say that either or  can be zero for the equation to be true.it's just like how in quadratics, when seeing  , the solutions are x=1 and x=2 because either one of those factors can be zero to satisfy the equation.if we focus on  , set it equal to zero and solve.    you need to use either unit circle knowledge or recall what the graph of cosine looks like.since cosine is defined to be, for a standard position angle  that has a point (x,y) on the terminal arm with r being the distance from the origin, we have    the places where we can get x = 0 are when the terminal arm of the angle is pointing directly up (pi/2) and when the terminal arm is pointing directly down (3pi/2). we cannot go past 2π due to the given restrictions on this question.so the solutions for  are    now, we have one other factor to consider:       set that equal to zero and solve.    from unit circle or the 30-60-90 triangle, we know that if we have a reference angle of  , we will get a sine ratio of  .sine is negative. sine is only negative in quadrants 3 and 4.from the unit circle,the angle in quadrant 3 with a sine ratio of  is  .the angle in quadrant 4 with a  sine ratio of    is  .there are no other solutions because we cannot go past 2π.therefore, the solutions to  are    bringing solutions from both factors, we have   

Do you need with the x value for the minimum? its always between the two intercepts for a parabola