Áp dụng BĐT Cauchy-Schwarz dạng Engel có:

\(A=\dfrac{1}{x^2+y^2}+\dfrac{1}{2xy}+\dfrac{1}{2xy}\ge\dfrac{4}{x^2+y^2+2xy}+\dfrac{1}{\dfrac{\left(x+y\right)^2}{2}}=\dfrac{4}{\left(x+y\right)^2}+\dfrac{2}{\left(x+y\right)^2}=6\)

Dấu "=" xảy ra khi x=y=\(\dfrac{1}{2}\)

a)

Đặt \(t=x-\dfrac{\pi}{6}\)

Pttt: \(sin2t=5sint+cos\left(3t-\dfrac{\pi}{2}\right)\)

\(\Leftrightarrow sin2t=5sint+sin3t\)

\(\Leftrightarrow sin2t-sin3t=5sint\)

\(\Leftrightarrow2.sint.cost+4sin^3t-3sint=5sint\)

\(\Leftrightarrow sint.cost+2sin^3t-4sint=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sint=0\\cost+2sin^2t-4=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}sint=0\\-2cos^2t+cost-2=0\left(vn\right)\end{matrix}\right.\)\(\Leftrightarrow t=k\pi\left(k\in Z\right)\)

\(\Rightarrow x=\dfrac{\pi}{6}+k\pi\left(k\in Z\right)\)

Vậy...

c,

Đặt \(t=x+\dfrac{\pi}{3}\)

Pttt: \(8cos^3t=cos\left(3t-\pi\right)\)

\(\Leftrightarrow8cos^3t=-cos3t\)\(\Leftrightarrow8cos^3t=3cost-4cos^3t\)

\(\Leftrightarrow4cos^3t-cost=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cost=0\\cost=\pm\dfrac{1}{2}\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}t=\dfrac{\pi}{2}+k\pi\\t=\pm\dfrac{\pi}{3}+k2\pi\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{6}+k\pi\\x=k2\pi\\x=-\dfrac{2\pi}{3}+k2\pi\end{matrix}\right.\)\(\left(k\in Z\right)\)

Vậy...

b) Pt \(\Leftrightarrow32.cos^6\left(x+\dfrac{\pi}{4}\right)-sin6x=1\)

\(\Leftrightarrow4\left[cos\left(2x+\dfrac{\pi}{2}\right)+1\right]^3-sin6x=1\)

\(\Leftrightarrow4\left(-sin2x+1\right)^3-sin6x=1\)

Đặt \(t=2x\)

Pttt \(4\left(-sint+1\right)^3-sin3t=1\)

\(\Leftrightarrow4\left(-sin^3t+3sin^2t-3sint+1\right)-\left(3sint-4sin^3t\right)=1\)

\(\Leftrightarrow12sin^2t-15sint+3=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sint=1\\sint=\dfrac{1}{4}\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{4}+k\pi\\x=\dfrac{1}{2}.arc.sin\left(\dfrac{1}{4}\right)+k\pi\\x=\dfrac{\pi}{2}-\dfrac{1}{2}.arc.sin\left(\dfrac{1}{4}\right)+k\pi\end{matrix}\right.\), k nguyên

Vậy...

d) Đặt \(t=x+\dfrac{\pi}{6}\)

\(\Rightarrow3x=3t-\dfrac{\pi}{2}\)

Pttt:\(2cost=sin\left(3t-\dfrac{\pi}{2}\right)-cos\left(3t-\dfrac{\pi}{2}\right)\)

\(\Leftrightarrow2cost=-cos3t-sin3t\)

\(\Leftrightarrow2cost=-3sint+4sin^3t+3cost-4cos^3t\)

\(\Leftrightarrow-4sin^3t+4cos^3t+3sint-cost=0\)

Thấy \(cost=0\) không là nghiệm pt

Tại \(cost\ne0\).Chia cả hai vế của pt cho \(cos^3t\) :

\(-4tan^3t+4+3tant.\dfrac{1}{cos^2t}-\dfrac{1}{cos^2t}=0\)

\(\Leftrightarrow-4tan^3t+4+3tant\left(1+tan^2t\right)-\left(1+tan^2t\right)=0\)

\(\Leftrightarrow-tan^3t-tan^2t+3tant+3=0\)

\(\Leftrightarrow\left[{}\begin{matrix}tant=-1\\tant=\pm\sqrt{3}\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}t=\dfrac{-\pi}{4}+k\pi\\t=\pm\dfrac{\pi}{3}+k\pi\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-5\pi}{12}+k\pi\\x=\dfrac{\pi}{6}+k\pi\\x=-\dfrac{\pi}{2}+k\pi\end{matrix}\right.\)\(\left(k\in Z\right)\)

Vậy...

b)Thấy sinx=0 không là nghiệm của pt

Chia cả hai vế của pt cho \(sin^3x\) ta được:

\(\dfrac{1}{sin^2x}-4+\dfrac{cosx}{sinx}.\dfrac{1}{sin^2x}=0\)

\(\Leftrightarrow1+cot^2x-4+cotx\left(1+cot^2x\right)=0\)

\(\Leftrightarrow cot^3x+cot^2x+cotx-3=0\)

\(\Leftrightarrow cotx=1\)\(\Leftrightarrow x=\dfrac{\pi}{4}+k\pi\left(k\in Z\right)\)

Vậy...

k) Pt \(\Leftrightarrow\left(sinx-cosx\right)\left(sin^2x+sinx.cosx+cos^2x\right)=sinx+cosx\)

\(\Leftrightarrow\left(sinx-cosx\right)\left(1+sinx.cosx\right)=sinx+cosx\)

\(\Leftrightarrow sin^2x.cosx-sinx.cos^2x-2cosx=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cosx=0\\sin^2x-sinx.cosx-2=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}cosx=0\\-sin^2x-sinx.cosx-2cos^2x=0\left(vn\right)\end{matrix}\right.\)

\(\Rightarrow x=\dfrac{\pi}{2}+k\pi\) (\(k\in Z\))

Vậy...

Còn nha,đang định làm nốt mấy bài nữa xong ngủ

a) \(2cosx.cos2x.cos3x-7=7cos2x\)

\(\Leftrightarrow\left(cos2x+cos4x\right).cos2x=7\left(cos2x+1\right)\)

\(\Leftrightarrow\left(cos2x+2cos^22x-1\right)cos2x=7\left(cos2x+1\right)\)

\(\Leftrightarrow\left(cos2x+1\right)\left(2cos2x-1\right)cos2x=7\left(1+cos2x\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}cos2x+1=0\\2cos^22x-cos2x=7\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}cos2x+1=0\\cos2x=\dfrac{1+\sqrt{57}}{4}\left(vn\right)\\cos2x=\dfrac{1-\sqrt{57}}{4}\left(vn\right)\end{matrix}\right.\)\(\Rightarrow2x=\pi+k2\pi,k\in Z\)

\(\Leftrightarrow x=\dfrac{\pi}{2}+k\pi\) ( k nguyên)

Vậy...

c) Pt\(\Leftrightarrow2sin^3x+2sin^2x-1+cosx=0\)

\(\Leftrightarrow2sin^2x\left(sinx+1\right)+\left(cosx-1\right)=0\)

\(\Leftrightarrow\left(1-cosx\right)\left(1+cosx\right)\left(2sinx+2\right)+\left(cosx-1\right)=0\)

\(\Leftrightarrow\left(1-cosx\right)\left(2sinx+2sinx.cosx+2cosx+1\right)=0\)

\(\Leftrightarrow\left(1-cosx\right)\left[2\left(sinx+cosx\right)+\left(sin^2x+2sinx.cosx+cos^2x\right)\right]=0\)

\(\Leftrightarrow\left(1-cosx\right)\left(sinx+cosx\right)\left(2+sinx+cosx\right)=0\) (I)

Có \(sinx+cosx=\sqrt{2}sin\left(x+\dfrac{\pi}{4}\right)\ge-\sqrt{2}\)

\(\Rightarrow2+sinx+cosx\ge2-\sqrt{2}>0\)

Từ (I)\(\Rightarrow\left[{}\begin{matrix}1-cosx=0\\sinx+cosx=0\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}cosx=1\\tanx=-1\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=k2\pi\\x=\dfrac{-\pi}{4}+k\pi\end{matrix}\right.\)\(\left(k\in Z\right)\)

Vậy...

e) Pt \(\Leftrightarrow8sinx.cosx-3\left(1-2sin^2x\right)=12sinx-3\)

\(\Leftrightarrow8sinx.cosx+6sin^2x=12sinx\)

\(\Leftrightarrow2sinx\left(4cosx+3sinx-6\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx=0\\4cosx+3sinx=6\left(vn\right)\end{matrix}\right.\)(vô nghiệm vì 4²+3²<6²)\(\Rightarrow x=k\pi\) \(\left(k\in Z\right)\)

Vậy...

a,

PT\(\Leftrightarrow sin^22x-cos^28x=sin\left(10x+\dfrac{\pi}{2}\right)\)

\(\Leftrightarrow sin^22x-cos^28x=cos10x\)

\(\Leftrightarrow\dfrac{1-cos4x}{2}-\dfrac{1+cos16x}{2}=cos10x\)

\(\Leftrightarrow-cos4x-cos16x=2cos10x\)

\(\Leftrightarrow-2.cos10x.cos6x=2cos10x\)

\(\Leftrightarrow cos10x+cos10x.cos6x=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cos10x=0\\1+cos6x=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}cos10x=0\\cos6x=-1\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{20}+\dfrac{k\pi}{10}\\x=\dfrac{\pi}{6}+\dfrac{k\pi}{3}\end{matrix}\right.\) , \(k\in Z\)

Vậy...

c) Pt\(\Leftrightarrow\dfrac{1+cos2x}{2}=cos\dfrac{4x}{3}\)

\(\Leftrightarrow1+cos2x-2cos\dfrac{4x}{3}=0\)

Đặt \(t=\dfrac{2x}{3}\). Pttt \(1+cos3t-2cos2t=0\)

\(\Leftrightarrow1+4cos^3t-3cost-2\left(2cos^2t-1\right)=0\)

\(\Leftrightarrow4cos^3t-4cos^2t-3cost+3=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cost=1\\cost=\dfrac{\sqrt{3}}{2}\\cost=-\dfrac{\sqrt{3}}{2}\end{matrix}\right.\)

TH1:\(cost=1\Leftrightarrow t=k2\pi\)\(\Leftrightarrow\dfrac{2x}{3}=k2\pi\)\(\Leftrightarrow x=k3\pi\) (\(k\in Z\))

TH2:\(cost=\dfrac{\sqrt{3}}{2}\Leftrightarrow t=\pm\dfrac{\pi}{6}+k2\pi\)\(\Leftrightarrow x=\pm\dfrac{\pi}{4}+k3\pi\)\(\left(k\in Z\right)\)

TH3:\(cost=-\dfrac{\sqrt{3}}{2}\Leftrightarrow t=\pm\dfrac{5\pi}{6}+k2\pi\)\(\Leftrightarrow x=\pm\dfrac{5\pi}{4}+k3\pi\)\(\left(k\in Z\right)\)

Vậy...

\(sin2x=sinx+1\)

\(\Rightarrow\left\{{}\begin{matrix}sin2x\ge0\\sin^22x=\left(sinx+1\right)^2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}sin2x\ge0\\4sin^2x.cos^2x=\left(sinx+1\right)^2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}sin2x\ge0\\4sin^2x\left(1-sin^2x\right)=\left(sinx+1\right)^2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}sin2x\ge0\\\left(sinx+1\right)\left(4sin^2x-4sin^3x-sinx-1\right)=0\end{matrix}\right.\)

Bấm máy thấy pt \(-4sin^3x+4sin^2x-sinx-1=0\) có một nghiệm \(sinx< 0\) không thỏa mãn \(sin2x\ge0\)

(Hoặc thử sd phương pháp cardano xem, chắc sẽ tìm được cụ thể nghiệm)

\(\Rightarrow sinx=-1\Leftrightarrow x=-\dfrac{\pi}{2}+k2\pi\) (\(k\in Z\))

Vậy...

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