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[TEX]4(\sin^2 x + \cos^4 x) = 5\cos x[/TEX]
[TEX]\Leftrightarrow 4[(\sin^2 x+\cos^2 x)^2 - 2\sin^2 x.\cos^2 x ] -5 \cos x = 0[/TEX]
[TEX]\Leftrightarrow 4[(1 - 2(1- \cos^2x ).\cos^2 x]-5\cos x =0[/TEX]
[TEX]4[1-2\cos^2 x + 2\cos^4 x] - 5\cos x=0[/TEX]
[TEX]8\cos^4x - 8\cos^2 x - 5\cos x + 4 =0[/TEX]
Đặt [TEX]\cos x = t; t\in [-1; 1][/TEX]
[TEX]8t^4 - 8t^2 - 5t +4 =0[/TEX]
[TEX]\Leftrightarrow [8t^4 - 2t^2] -[6t^2+5t-4]=0[/TEX]
[TEX]\Leftrightarrow 2t^2(2t-1)(2t+1) - (2t-1)(t+1) = 0[/TEX]
[TEX](2t-1)(4t^3 +2t^2 -3t -4)=0[/TEX]
[TEX]\left[\begin{2t-1 =0 }\\{4t^3 +2t^2 -3t -4=0} [/TEX]
Với [TEX]4t^3 +2t^2 -3t -4 = 0[/TEX]
[TEX]\Leftrightarrow 4\cos^3 +2\cos^2 x-3\cos x-4 =0[/TEX]
[TEX]\Leftrightarrow (4\cos^3 x -3 \cos x) + (2\cos^2 x -1) - 3 =0[/TEX]
[TEX]\Leftrightarrow \cos 3x + \cos 2x -3 =0[/TEX]
Vô nghiệm do [TEX]\cos 3x \leq 1; \cos 2x \leq 1[/TEX] nên [TEX]\cos 3x + \cos 2x -3 \leq -1 < 0[/TEX]
Với [TEX]2t=1 \Leftrightarrow t= \frac{1}{2} \Leftrightarrow \cos x = \frac{1}{2} \Leftrightarrow x=\pm \frac{\pi}{3} + k2\pi[/TEX]
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