The general solution of 4 sin^2(x) - 4 sin(x) | vedclass

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Question

(C) Given the equation: $4 \sin^2(x) - 4 \sin(x) + 1 = 0$ This is a quadratic equation in terms of $\sin(x)$,which can be written as $(2 \sin(x) - 1)^

Vedclass · Accepted Answer

$x = n\pi + (-1)^n \frac{\pi}{6}, n \in \mathbb{Z}$

Vedclass · Answer

(C) Given the equation: $4 \sin^2(x) - 4 \sin(x) + 1 = 0$ This is a quadratic equation in terms of $\sin(x)$,which can be written as $(2 \sin(x) - 1)^2 = 0$. Taking the square root on both sides,we get: $2 \sin(x) - 1 = 0$ $\sin(x) = \frac{1}{2}$ We know that $\sin(x) = \sin(\frac{\pi}{6})$. The general solution for $\sin(x) = \sin(\alpha)$ is $x = n\pi + (-1)^n \alpha$,where $n \in \mathbb{Z}$. Therefore,$x = n\pi + (-1)^n \frac{\pi}{6}, n \in \mathbb{Z}$. Hence,option $C$ is correct.

The general solution of $4 \sin^2(x) - 4 \sin(x) + 1 = 0$ is

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