Question

Interval is 0, 2pi) 4 sin^2x-4 sin x + 1 = 0

Answer 1

Answer:

$\large\boxed{x=\dfrac{\pi}{6}\ \vee\ x=\dfrac{5\pi}{6}}$

Step-by-step explanation:

$4\sin^2x-4\sin x+1=0\\\\2^2\sin^2x-2(2\sin x)(1)+1^2=0\\\\(2\sin x)^2-2(2\sin x)(1)+1^2=0\qquad\text{use}\ (a-b)^2=a^2-2ab+b^2\\\\(2\sin x-1)^2=0\iff2\ain x-1=0\qquad\text{add 1 to both sides}\\\\2\sin x=1\qquad\text{divide both sides by 2}\\\\\sin x=\dfrac{1}{2}\to x=\dfrac{\pi}{6}\ \vee\ x=\dfrac{5\pi}{6}$