Algebra and Trig Equation

Your short question is interesting because it uses some important
algebra and trig ideas.

Probably the easiest approach is to start with the trig identity for
the cosine of 2 angles added together:

cos(A+B) = cos(A)*cos(B) - sin(A)*sin(B)

The particular version of this that you need here is when A = B:
cos(2A) = cos(A+A) = cos(A)*cos(A) - sin(A)*sin(A)

You may know the identity "sine squared + cosine squared = 1" or:
sin(A)*sin(A) + cos(A)*cos(A) = 1

Using both of these last 2 trig identities together you get:
cos(2A) = 1 - 2*sin(A)*sin(A)

Now, using this final identity, the original equation becomes:
1 = sin(3x) -(1 - 2*sin(3x)*sin(3x))

because 6x is 2 times 3x. This simplifies to:
2*sin(3x)*sin(3x) + sin(3x) - 2 = 0

Letting S stand for sin(3x), this is clearly a quadratic equation:
2*S*S + S - 2 = 0

Solve this for S with the quadratic formula, which gives 2 different
values for S. Because S = sin(3x) is a sine value it must be between
-1 and +1 so only one of the solutions can be used. It is:
S = sin(3x) = (sqrt(17)-1)/4

Taking the inverse sine of both sides gives:
3x = arcsin[(sqrt(17)-1)/4]
x = arcsin[(sqrt(17)-1)/4]/3

In other words, x is 1/3 of the angle whose sine is (sqrt(17)-1)/4,
or about 17.11057 degrees. I will leave it for you to use this x
value in the original equation to verify that it is the solution.

I hope this is what you need. Best regards.