Question

9sin^2x-18sin+9=0

Answer 1

Your question is: \(\large\color{slate}{\displaystyle 9\sin^2(x)-18\sin(x)+9=0 }\)

perfect square trinomial (if you think of sin(x) as of a.)

Answer 2

you can say `let sin(x)=a` and then factor it, if you can't factor it right away.

Answer 3

9(x-1)^2

Answer 4

Yes. that is the form, but it is sin(x) or "a", but not just x.

Answer 5

Well, if sin(x) = a: 9(a-1)^2 = 0

Answer 6

But that is the correct factorization, yes.

Answer 7

But you can go a few steps further...

Answer 8

find all solutions to the equation.

Answer 9

\(\large\color{slate}{ 9\sin^2x-18\sin(x)+9=0 }\)

\(\large\color{slate}{ 9(\sin (x)-1)^2=0 }\)

Answer 10

divide both sides by 9, and take the square root of both sides.

Answer 11

pi/2