Question

4 sin^2 x - 4 sin x + 1 = 0 find all solutions in the interval [0,2pi]

Answer 1

Hello.

$\mathsf{\boxed{4 \sin^{2} x - 4 sin x + 1 = 0}}$

In order to ease comprehension, let's replace 'sin x' with 'y'.

$\mathsf{4y^{2} - 4y + 1 = 0}$

$\mathsf{\triangle = b^{2} - 4ac} \\ \\ \mathsf{\triangle = 16 - 4 \times 4 \times 1} \\ \\ \mathsf{\triangle = 0}$

$\mathsf{y = \dfrac{-b \pm \sqrt{\triangle}}{2a}} \\ \\ \mathsf{y = \dfrac{4 \pm 0}{8}} \\ \\ \mathsf{y = \dfrac{1}{2}} \\ \\ \therefore \\ \\ \sin x = \dfrac{1}{2}$

In the interval: [0, 2pi]:

y = pi/6 or 5pi/6

Hope I helped.

Answer 2

$4 \sin^2 x - 4 \sin x + 1 = 0\\ (2\sin x-1)^2=0\\ 2\sin x-1=0\\ 2\sin x=1\\ \sin x=\dfrac{1}{2}\\ x=\dfrac{\pi}{6} \vee x=\dfrac{5\pi}{6}$