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ivolga24 [154]
4 years ago
5

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

Mathematics
2 answers:
levacccp [35]4 years ago
7 0
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.
FrozenT [24]4 years ago
4 0
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}

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