87.
\(\left(sinx+\sqrt{3}cosx\right)sin3x=2\)
\(\Leftrightarrow\left(\dfrac{1}{2}sinx+\dfrac{\sqrt{3}}{2}cosx\right)sin3x=1\)
\(\Leftrightarrow sin\left(x+\dfrac{\pi}{3}\right).sin3x=1\)
\(\Leftrightarrow\dfrac{1}{2}cos\left(2x-\dfrac{\pi}{3}\right)-\dfrac{1}{2}cos\left(4x+\dfrac{\pi}{3}\right)=1\)
\(\Leftrightarrow cos\left(2x-\dfrac{\pi}{3}\right)-cos\left(4x+\dfrac{\pi}{3}\right)=2\)
Ta có:
\(cos\left(2x-\dfrac{\pi}{3}\right)\le1;cos\left(4x+\dfrac{\pi}{3}\right)\ge-1\)
\(\Rightarrow cos\left(2x-\dfrac{\pi}{3}\right)-cos\left(4x+\dfrac{\pi}{3}\right)\le2\)
Đẳng thức xảy ra khi:
\(\left\{{}\begin{matrix}cos\left(2x-\dfrac{\pi}{3}\right)=1\\cos\left(4x+\dfrac{\pi}{3}\right)=-1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x-\dfrac{\pi}{3}=k2\pi\\4x+\dfrac{\pi}{3}=\pi+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{\pi}{6}+k\pi\\x=\dfrac{\pi}{6}+\dfrac{k\pi}{2}\end{matrix}\right.\)
\(\Leftrightarrow x=\dfrac{\pi}{6}+k\pi\)
Câu 88:
\(\sqrt{5+sin^2x}=sinx+2cosx\)
\(\Leftrightarrow\sqrt{1+\dfrac{sin^2x}{5}}=\dfrac{1}{\sqrt{5}}sinx+\dfrac{2}{\sqrt{5}}cosx\)
\(\Leftrightarrow\sqrt{1+\dfrac{sin^2x}{5}}=sin\left(x+arc.cos\dfrac{1}{\sqrt{5}}\right)\)
Có \(\sqrt{1+\dfrac{sin^2x}{5}}\ge\sqrt{1+0}=1\)
\(sin\left(x+arc.cos\dfrac{1}{\sqrt{5}}\right)\le1\)
Dấu "=" xảy ra khi \(\Leftrightarrow\left\{{}\begin{matrix}sin^2x=0\\sin\left(x+arc.cos\dfrac{1}{\sqrt{5}}\right)=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=k\pi\\x+arc.cos\dfrac{1}{\sqrt{5}}=k\pi\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=k\pi\\x=k\pi-arc.cos\dfrac{1}{\sqrt{5}}\end{matrix}\right.\)\(\Rightarrow x\in\varnothing\)
Vậy \(S=\varnothing\)
Câu 89:
\(cos^24x+cos^28x=sin^212x+sin^216x+2\)
\(\Leftrightarrow cos^24x+cos^28x=sin^212x+sin^216x+sin^24x+cos^24x+sin^28x+cos^28x\)
\(\Leftrightarrow sin^212x+sin^216x+sin^24x+sin^28x=0\)
Có \(VT\ge0\)
Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}sin12x=0\\sin16x=0\\sin4x=0\\sin8x=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{k\pi}{12}\\x=\dfrac{k\pi}{16}\\x=\dfrac{k\pi}{4}\\x=\dfrac{k\pi}{8}\end{matrix}\right.\)\(\Rightarrow x=0\)
Vậy x=0
88.
\(\sqrt{5+sin^2x}=sinx+2cosx\)
Ta có:
\(VP=sinx+2cosx\le\sqrt{\left(1^2+2^2\right)\left(sin^2x+cos^2x\right)}=\sqrt{5}\)
\(VT=\sqrt{5+sin^2x}\ge\sqrt{5}\)
\(\Rightarrow\sqrt{5+sin^2x}\ge sinx+2cosx\)
Đẳng thức xảy ra khi \(\left\{{}\begin{matrix}sinx+2cosx=\sqrt{5}\\sinx=0\end{matrix}\right.\left(vn\right)\)
Vậy phương trình đã cho vô nghiệm.
\(cos^24x+cos^28x=sin^212x+sin^216x+2\)
\(\Leftrightarrow2-sin^24x-sin^28x=sin^212x+sin^216x+2\)
\(\Leftrightarrow sin^24x+sin^28x+sin^212x+sin^216x=0\)
\(\Leftrightarrow sin4x=sin8x=sin12x=sin16x=0\)
\(\Leftrightarrow x=\dfrac{k\pi}{4}\)
