General solution of the equation 4cosx−3secx=tanx,(cosx≠0) can be
A
nπ+(−1)nα,sinα=−1+√178,n∈Z
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B
nπ+(−1)nβ,sinβ=−1−√178,n∈Z
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C
nπ+(−1)nα,sinα=1+√178,n∈Z
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D
nπ+(−1)nβ,sinβ=1−√178,n∈Z
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Solution
The correct option is Bnπ+(−1)nβ,sinβ=−1−√178,n∈Z 4cosx−3secx=tanx⇒4cosx−3cosx=sinxcosx⇒4cos2x−3=sinx⇒4(1−sin2x)−3=sinx⇒4sin2x+sinx−1=0⇒sinx=−1±√178
If sinx=−1+√178 ∴x=nπ+(−1)nα,sinα=−1+√178,n∈Z
If sinx=−1−√178 ∴x=nπ+(−1)nβ,sinβ=−1−√178,n∈Z