Question

HELP!! Medal will be given!

Answer 1

Find all solutions in the interval [0, 2π).

4 sin^2x - 4 sin x + 1 = 0

Answer 2

A good starting point is to factor the equation as much as possible:
\[4\sin^2x-4\sin x+1=0\Leftrightarrow 4\sin x(\sin x-1)+1=0\]
Now it's still pretty hard to 'see' the solution, but we could rephrase the question as to find the zeroes of the polynomial \(4q(q-1)+1=0\), but with \(q\) in place of \(\sin x\). That is, we substitute \(q=\sin x\). If you find the roots to this, you can either find the angle directly if you have memorized sin for certain angles, or you can reverse the substitution \(x=\arcsin q\) and punch the roots into that on your calculator to find the solution(s).