Determine all values of ′a′ for which the equation cos4x−(a+2)cos2x−(a+3)=0, possesses solution.Find the solutions.
A
x=nπ±sin−1√a+3,aϵ[−3,−2];nϵz
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B
x=2nπ±cos−1√a+3,aϵ[−3,−2];nϵz
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C
x=nπ±cos−1(a+3),aϵ[−3,−2];nϵz
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D
x=nπ±cos−1√a+3,aϵ[−3,−2];nϵz
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Solution
The correct option is Dx=nπ±cos−1√a+3,aϵ[−3,−2];nϵz cos4x−(a+2)cos2x−(a+3)=0 cos2x=a+2±√(a+2)2+4(a+3)2=a+2±√(a+4)22 =a+2+a+42,a+2−a−42=a+3,0 cos2x=a+3⇒x=nπ±cos−1√a+3 But cos−1√a+3 is defined when −1≤a+3≤+1