AP EAMCET Engineering 18 Sep 2020 Shift 2 Paper

Section: Mathematics

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Question : 22 of 160

Marks: +1, -0

General solution of 4 sin2(x)−4‌sin(x)+1=0 is

x=2nπ±
π
3
,n∊Z
x=nπ+(−1)n
π
3
, n∊Z
x=nπ+(−1)n
π
6
, n∊Z
x=nπ+(−1)n
π
4
, n∊Z

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