I don't know about you, but I definitely don't like to solve equations involving #sin^2x#. It looks really messy. To make it a little neater, let's use the substitution #u=sinx#. Our equation becomes:
#6u^2-u-2=0#
Ah...that's better. Now we have a quadratic equation that we know how to solve. Let's try to factor it, using the AC method. First, we multiply the #color(white)(x)^2# term by the constant term - in this case, the #6# in our equation by the #-2#:
#6*-2=-12#
Then, we find two numbers that multiply to #-12# and add to the middle term, in this case the #-1#. We'll need to list the factors of #-12#:
#color(white)(XI)ul("Factors of -12")#
#-1*12# or #1*-12#
#-2*6# or #2*-6#
#-3*4# or #3*-4#
Which of these adds to #-1#? Take a good look and you'll see #3*-4=-12# and #3-4=-1#. These numbers satisfy our conditions. The next step is breaking #-u# into #3u-4u#, which are the numbers that add to #-1#:
#6u^2+3u-4u-2=0#
We can factor out a #3u# from the first pair and a #-2# from the second pair:
#3ucolor(red)((2u+1))-2color(red)((2u+1))=0#
Note the common factor #2u+1#; we can pull that out too:
#color(red)((2u+1))(3u-2)=0#
Using the zero product property, we can set our two factors equal to #0# and solve:
#2u+1=0# and #3u-2=0#
#->u=-1/2# and #u=2/3#
Remember that #u=sinx#, so:
#sinx=-1/2# and #sinx=2/3#
Now we just have to solve this on #[0,2pi]#. So, when does #sinx# equal #-1/2#? The unit circle tells us that #sinx=-1/2# when #x=(7pi)/6# and #x=(11pi)/6#, so those are two solutions.
How about #sinx=2/3#? We'll need to use a calculator for that. Take the inverse sine of both sides:
#sin^(-1)(sinx)=sin^(-1)(2/3)#
#->x=sin^(-1)(2/3)#
Using a calculator, we obtain the principal solution #x~~0.73#. However, there is another solution that the calculator won't give us - the solution in the second quadrant. To find that, we subtract #0.73# from #pi#:
#x=pi-0.73~~2.41#
Thus our solutions are #x=(7pi)/6#, #x=(11pi)/6#, #x=0.73#, and #x=2.41#.
