Question

2sinAcosA = squareroot of 3/2

find the general solution in terms of 360 degrees * n or 2pin
find the particular values of x in the domain 0 degrees lessthan/equal to thetha lessthan 360degrees
i need help please

Answer 1

Well, the first thing to realize is that 2sinAcosA = sin(2A). So, your new equation is:

\[sin{2A} = \frac{\sqrt{3}}{2}\]

We know that sin(pi/3) and sin(2pi/3) are both sqrt3/2. That means two solutions for A are pi/6 and pi/3 (just dividing those by two). Also, sine has a period of 2pi, and we divided by two, so both of those solutions recur with period pi. So your answer is pi/6+pi*n and pi/3+pi*n.

Answer 2

how come it isnt 2pi?