vivykalynn
vivykalynn vivykalynn
  • 23-04-2019
  • Mathematics
contestada

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

Respuesta :

gmany
gmany gmany
  • 23-04-2019

Answer:

[tex]\large\boxed{x=\dfrac{\pi}{6}\ \vee\ x=\dfrac{5\pi}{6}}[/tex]

Step-by-step explanation:

[tex]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}[/tex]

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