How do you solve 3sin2AtanA=2 for 0<A<180?

The first step in solving any trigonometric equation is to convert everything into sine and cosine functions. So in this case we need to start by converting the tanA. We can use the identity: tanA=sinA/cosA

Hence the equation becomes:

3sin2A x sinA/cosA=2

Next we need to deal with the double angle that we have in sin2A. This time we can use the double angle rule for sine which says: sin2A=2sinAcosA. We substitute this into our equation to get:

3 x 2sinAcosA x sinA/cosA=2

We can now simplify the equation by cancelling out cosA and combining sinA:

6sin^​2A=2

which we simplify further to:

sin^​2​A=1/3

square rooting both sides we get:

sinA=0.5774

hence by taking the inverse of sine we get:

A=35.27 degrees

However this is only one possible solution. We need to use the general solution of sine to find all possible answers within the given angle range.

The general solution for sine is:

sin^-1​=A +360k or 180-A +360k where k is an integer

Hence our answer becomes:

A= (35.27 or  144.73) +360k

but since only 35.27 and 144.73 lie in the given range of 0<A<180 our possible answers are just:

A=35.27 or 144.73