I don't know about you, but I definitely don't like to solve equations involving sin^2xsin2x. It looks really messy. To make it a little neater, let's use the substitution u=sinxu=sinx. Our equation becomes:
6u^2-u-2=06u2−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)^2x2 term by the constant term - in this case, the 66 in our equation by the -2−2:
6*-2=-126⋅−2=−12
Then, we find two numbers that multiply to -12−12 and add to the middle term, in this case the -1−1. We'll need to list the factors of -12−12:
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.
