ANSWER
[tex]x = \frac{\pi}{6} \: or \: x = \frac{5\pi}{6} [/tex]
EXPLANATION
The given trigonometric equation is
[tex]4 \sin ^{2} x - 4 \sin(x) + 1 = 0[/tex]
This is a quadratic equation in sinx.
We split the middle term to obtain,
[tex]4 \sin ^{2} x - 2 \sin(x) - 2 \sin(x) + 1 = 0[/tex]
Factor by grouping to get,
[tex]2 \sin(x) (2 \sin(x) - 1) - 1(2 \sin(x) - 1) = 0[/tex]
This implies that,
[tex](2 \sin(x) - 1)(2 \sin(x) - 1) = 0[/tex]
[tex] \sin(x) = 0.5[/tex]
This gives us,
[tex]x = \frac{\pi}{6} [/tex]
in the first quadrant.
Or
[tex]x = \pi - \frac{\pi}{6} [/tex]
[tex]x = \frac{5\pi}{6} [/tex]
in the second quadrant.
Answer:
x = [tex]\frac{\pi}{6}[/tex], [tex]\frac{7\pi}{6}[/tex], [tex]\frac{11\pi}{6}[/tex].
Step-by-step explanation:
The given equation is 4 sin²x - 4 sin x + 1 = 0
(2sinx)² - 2(2sinx) + 1 = 0
(2sinx - 1 )² = 0
Sinx = [tex]\frac{1}{2}[/tex] ⇒ x = sin⁻¹ ( [tex]\frac{1}{2}[/tex])
So between the interval [0, 2π] value of x will be [tex]\frac{\pi}{6}[/tex], [tex]\frac{7\pi}{6}[/tex], [tex]\frac{11\pi}{6}[/tex]
[Since sine is positive in 1st 3rd and 4th quadrant]
So value of x will be x = [tex]\frac{\pi}{6}[/tex], [tex]\frac{7\pi}{6}[/tex], [tex]\frac{11\pi}{6}[/tex].
