General solution of the equation 2sin2x+3cot2x−4sinx−6cotx+5=0 is
A
x=(4n+1)π2wheren∈Z
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B
x=(4n+1)π4wheren∈Z
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C
x=(2n+1)π2wheren∈Z
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D
x=ϕ
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Solution
The correct option is Dx=ϕ 2sin2x+3cot2x−4sinx−6cotx+5=0 ⇒2sin2x−4sinx+2+3cot2x−6cotx+3=0 ⇒2(sinx−1)2+3(cotx−1)2=0 ⇒sinx−1=0 and cotx−1=0 ⇒sinx=1 and cotx=1 But when sinx=1⇒cotx=0 as cotx=cosxsinx So, x=ϕ