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4cos^2x+sin^2x=?
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\[4\frac{ 1+\cos 2x }{2 }+\frac{ 1-\cos 2x }{2 }=0\] 4+4cos 2x+1-cos 2x=0 3 cos 2x=-5 cos 2x=-5/3 impossible as \[\left| \cos 2x \right|\le 1\]
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