General solution of \(4 \sin ^2(x)-4 \sin (x)+1=0\) is
General solution of \(4 \sin ^2(x)-4 \sin (x)+1=0\) is
- \(x=2 n \pi \pm \frac{\pi}{3}, n \in Z\)
- \(x=n \pi+(-1)^n \frac{\pi}{3}, n \in Z\)
- \(x=n \pi+(-1)^n \frac{\pi}{6}, n \in Z\)
- \(x=n \pi+(-1)^n \frac{\pi}{4}, n \in Z\)
Solution
\(\begin{aligned}
4 \operatorname{Sin}^2 x-4 \operatorname{Sin} x+1 & =0 \\
(2 \operatorname{Sin} x-1)^2 & =0 \\
2 \operatorname{Sin} x-1 & =0 \\
\operatorname{Sin} x & =\frac{1}{2} \\
x & =n \pi+(-1)^n \cdot \frac{\pi}{6}, n \in \mathbf{Z}
\end{aligned}\)
Hence, option (c) is correct.
Asked in: AP EAMCET 2020 (18 Sep Shift 2)
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