Question

Find all solutions in the interval [0, 2π). 4 sin2 x - 4 sin x + 1 = 0

Answer 1

ANSWER

$x = \frac{\pi}{6} \: or \: x = \frac{5\pi}{6}$

EXPLANATION

The given trigonometric equation is

$4 \sin ^{2} x - 4 \sin(x) + 1 = 0$

This is a quadratic equation in sinx.

We split the middle term to obtain,

$4 \sin ^{2} x - 2 \sin(x) - 2 \sin(x) + 1 = 0$

Factor by grouping to get,

$2 \sin(x) (2 \sin(x) - 1) - 1(2 \sin(x) - 1) = 0$

This implies that,

$(2 \sin(x) - 1)(2 \sin(x) - 1) = 0$

$\sin(x) = 0.5$

This gives us,

$x = \frac{\pi}{6}$

in the first quadrant.

Or

$x = \pi - \frac{\pi}{6}$

$x = \frac{5\pi}{6}$

in the second quadrant.

Answer 2

Answer:

x = $\frac{\pi}{6}$, $\frac{7\pi}{6}$,  $\frac{11\pi}{6}$.

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 = $\frac{1}{2}$ ⇒ x = sin⁻¹ ( $\frac{1}{2}$)

So between the interval [0, 2π] value of x will be  $\frac{\pi}{6}$,  $\frac{7\pi}{6}$,  $\frac{11\pi}{6}$

[Since sine is positive in 1st 3rd and 4th quadrant]

So value of x will be x =  $\frac{\pi}{6}$, $\frac{7\pi}{6}$,  $\frac{11\pi}{6}$.