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
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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 \(cos^3t\) :
\(-4tan^3t+4+3tant.\dfrac{1}{cos^2t}-\dfrac{1}{cos^2t}=0\)
\(\Leftrightarrow-4tan^3t+4+3tant\left(1+tan^2t\right)-\left(1+tan^2t\right)=0\)
\(\Leftrightarrow-tan^3t-tan^2t+3tant+3=0\)
\(\Leftrightarrow\left[{}\begin{matrix}tant=-1\\tant=\pm\sqrt{3}\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}t=\dfrac{-\pi}{4}+k\pi\\t=\pm\dfrac{\pi}{3}+k\pi\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-5\pi}{12}+k\pi\\x=\dfrac{\pi}{6}+k\pi\\x=-\dfrac{\pi}{2}+k\pi\end{matrix}\right.\)\(\left(k\in Z\right)\)
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