The complete solution of the equation 7cos2x+sinxcosx−3=0 is given by
A
nπ+π2,(n∈I)
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B
nπ−π4,(n∈I)
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C
nπ−tan−143,(n∈I)
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D
nπ+tan−143,(n∈I)
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Solution
The correct option is Cnπ−π4,(n∈I) 7cos2x+sinxcosx−3=0⇒1+7cos2x+sin2x=0 Substituting y=tanx⇒sin2x=2y1+y2,cos2x=1−y21+y2 ⇒−2(3y2−y−4)y2+1=0⇒3y2−y−4=0⇒(y+1)(3y−4)=0 ⇒y=−1 or 3y−4=0 ⇒tanx=−1 or tanx=43 ⇒x=πn−π4 or x=πm+tan−143