General solution of the equation sin2x+cosec2x−2(sinx+cosecx)+2=0 is
A
(4n−1)π2,n∈Z
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B
(2n−1)π2,n∈Z
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C
(4n+1)π2,n∈Z
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D
(2n+1)π2,n∈Z
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Solution
The correct option is C(4n+1)π2,n∈Z sin2x+cosec2x−2(sinx+cosecx)+2=0 ⇒(sin2x−2sinx+1)+(cosec2x−2cosecx+1)=0 ⇒(sinx−1)2+(cosecx−1)2=0 ⇒sinx−1=0 and cosecx−1=0 ⇒sinx=1 and cosecx=1 ⇒x=(4n+1)π2;wheren∈Z