1. Tìm Min $y = 3\sin^2 x+4\sin x\cos x-5\cos^2 x+2$
2. Tìm Max $y = \dfrac{16}{3}(\sin^3 x\cos 3x+\cos^3 x\sin 3x)+3\cos 4x$
1.Ta có; $y=5sin^{2}x-5cos^{2}x+2cos^{2}x+4sinx.cosx= 2sin2x-4cos2x-1$
2.Ta có: $y=\frac{16}{3}(sin^{3}x(4cos^{3}x-3cosx)+cos^{3}x(3sinx-4sin^{3}x))+3cos4x=4sin4x+3cos4x$
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