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Questions  

If 2cosA2=1+sinA+1sinA, then A2

a
2nπ+π4 and 2nπ+3π4
b
2nπ−π4 and 2nπ+π4
c
2nπ−3π4 and 2nπ−π4
d
−∞ and +∞

detailed solution

Correct option is B

We have,2cos⁡A2=1+sin⁡A+1−sin⁡A⇒2cos⁡A2=cos⁡A2+sin⁡A22+cos⁡A2−sin⁡A22⇒2cos⁡A2=cos⁡A2+sin⁡A2+cos⁡A2−sin⁡A2⇒cos⁡A2+sin⁡A2≥0 and cos⁡A2−sin⁡A2≥0⇒−3π4≤A2≤3π4 and −π4≤A2≤π4⇒−π4≤A2≤π4⇒ 2nπ−π4≤A2≤2nπ+π4,n∈Z

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