If 3cosx≠2sinx, then the general solution of sin2x−cos2x=2−sin2x is
A
nπ+(−1)nπ2,n∈Z
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B
nπ2,n∈Z
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C
(4n±1)π2,n∈Z
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D
(2n−1)π,n∈Z
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Solution
The correct option is C(4n±1)π2,n∈Z Given equation is sin2x–cos2x=2–sin2x ⇒1–cos2x–(2cos2x–1)=2–2sinxcosx ⇒cosx(2sinx–3cosx)=0 ⇒cosx=0(∵2sinx−3cosx≠0) ⇒x=4n±1)π2