The general solution satisfying the trigonometric equations 4sin3x+1=sinx(1+4sinx) and cos3x=3–4cos2x is given by
A
x=nπ±π6, only n∈Z
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B
x=nπ+(−1)nπ3 only, n∈Z
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C
x=2nπ±π6 only n∈Z
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D
x=nπ+(−)nπ6, n∈Z
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Solution
The correct option is Ax=nπ±π6, only n∈Z 4sin3x−4sin2x−sinx+1=0 (4sin2x−1)(sinx−1)=0 ⇒sinx=±12,sinx=1 ... (i) 4cos3x−3cosx=3−4cos2x ⇒4cos3x+4cos2x−3cosx−3=0 (4cos2x−3)(cosx+1)=0 cosx=±√32,cosx=−1 ... (ii)
From (i) and (ii) x=nπ±π6