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c.
\(\Leftrightarrow\frac{1}{2}-\frac{1}{2}cos\left(8x+\frac{2\pi}{3}\right)=\frac{1}{2}-\frac{1}{2}cos\left(\frac{14\pi}{5}-2x\right)\)
\(\Leftrightarrow cos\left(8x+\frac{2\pi}{3}\right)=cos\left(2\pi+\frac{4\pi}{5}-2x\right)\)
\(\Leftrightarrow cos\left(8x+\frac{2\pi}{3}\right)=cos\left(\frac{4\pi}{5}-2x\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}8x+\frac{2\pi}{3}=\frac{4\pi}{5}-2x+k2\pi\\8x+\frac{2\pi}{3}=2x-\frac{4\pi}{5}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{75}+\frac{k\pi}{5}\\x=-\frac{11\pi}{45}+\frac{k\pi}{3}\end{matrix}\right.\)
a.
\(\Leftrightarrow\frac{1}{2}+\frac{1}{2}cos4x=\frac{1}{2}-\frac{1}{2}cos\left(2x+\frac{2\pi}{3}\right)\)
\(\Leftrightarrow cos4x=-cos\left(2x+\frac{2\pi}{3}\right)\)
\(\Leftrightarrow cos4x=cos\left(\frac{\pi}{3}-2x\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}4x=\frac{\pi}{3}-2x+k2\pi\\4x=2x-\frac{\pi}{3}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{18}+\frac{k\pi}{3}\\x=-\frac{\pi}{6}+k\pi\end{matrix}\right.\)
b.
\(\Leftrightarrow\frac{1}{2}-\frac{1}{2}cos\left(10x+\frac{2\pi}{3}\right)-\frac{1}{2}-\frac{1}{2}cos\left(6x+\frac{\pi}{2}\right)=0\)
\(\Leftrightarrow cos\left(10x+\frac{2\pi}{3}\right)=-cos\left(6x+\frac{\pi}{2}\right)\)
\(\Leftrightarrow cos\left(10x+\frac{2\pi}{3}\right)=cos\left(\frac{\pi}{2}-6x\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}10x+\frac{2\pi}{3}=\frac{\pi}{2}-6x+k2\pi\\10x+\frac{2\pi}{3}=6x-\frac{\pi}{2}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{\pi}{96}+\frac{k\pi}{8}\\x=-\frac{7\pi}{24}+\frac{k\pi}{2}\end{matrix}\right.\)
Câu 1:
\(cos7x-\sqrt{3}sin7x=-2\\ \Leftrightarrow cos\left(7x+\dfrac{\pi}{3}\right)=-1\\ \Leftrightarrow7x+\dfrac{\pi}{3}=-\pi+k2\pi\\ \Leftrightarrow x=-\dfrac{4\pi}{21}+k\dfrac{2\pi}{7}\)
Vì \(x\in[\dfrac{2\pi}{5};\dfrac{6\pi}{7}]\)
\(\Rightarrow\dfrac{2\pi}{5}\le x\le\dfrac{6\pi}{7}\\ \Leftrightarrow\dfrac{2\pi}{5}\le-\dfrac{4\pi}{21}+k\dfrac{2\pi}{7}\le\dfrac{6\pi}{7}\\ \Leftrightarrow\dfrac{31}{15}\le k\le\dfrac{11}{3}\)
Vì \(k\in Z\) nên \(k=3\)
Vậy \(x\) cần tìm là \(\dfrac{2\pi}{3}\)
Câu 2:
\(2sin^2x-sinxcosx-cos^2x=m\\ \Leftrightarrow2\dfrac{1-cos2x}{2}-\dfrac{1}{2}s\text{in2}x-\dfrac{1+cos2x}{2}=m\\ \Leftrightarrow3cos2x+s\text{in2}x=1-2m\)
Điều kiện để phương trình có nghiệm là:
\(3^2+1^2\ge\left(1-2m\right)^2\\ \Leftrightarrow4m^2-4m-9\le0\\ \Leftrightarrow\dfrac{1-\sqrt{10}}{2}\le m\le\dfrac{1+\sqrt{10}}{2}\)
1) đặc : \(f\left(x\right)=y=cot4x\)
điều kiện xác định : \(sin4x\ne0\Leftrightarrow4x\ne k\pi\Leftrightarrow x\ne\dfrac{k\pi}{4}\)
\(\Rightarrow x\in D\) thì \(-x\in D\)
ta có : \(f\left(-x\right)=cot\left(-4x\right)=-cot4x=-f\left(x\right)\)
\(\Rightarrow\) hàm này là hàm lẽ
2) đặc : \(f\left(x\right)=y=\left|cotx\right|\)
điều kiện xác định : \(sinx\ne0\Leftrightarrow x\ne k\pi\)
\(\Rightarrow x\in D\) thì \(-x\in D\)
ta có : \(f\left(-x\right)=\left|cot\left(-x\right)\right|=\left|-cotx\right|=\left|cotx\right|=f\left(x\right)\)
\(\Rightarrow\) hàm này là hàm chẳn
3) đặc : \(f\left(x\right)=y=1-sin^2x=cos^2x\)
điều kiện xác định : \(D=R\)
\(\Rightarrow x\in D\) thì \(-x\in D\)
ta có : \(f\left(-x\right)=cos^2\left(-x\right)=cos^2x=f\left(x\right)\)
\(\Rightarrow\) hàm này là hàm chẳn
4) đặc : \(f\left(x\right)=y=sin\left(x+\dfrac{\pi}{4}\right)=\dfrac{sinx+cosx}{\sqrt{2}}\)
điều kiện xác định : \(D=R\)
\(\Rightarrow x\in D\) thì \(-x\in D\)
ta có : \(f\left(-x\right)=\dfrac{sin\left(-x\right)+cos\left(-x\right)}{\sqrt{2}}=\dfrac{-sinx+cosx}{\sqrt{2}}\ne f\left(x\right);-f\left(x\right)\)
\(\Rightarrow\) hàm này là hàm không chẳn không lẽ
mấy bài còn lại bn làm tương tự cho quen nha
ta có : \(sin^3x.cosx-cos^3x.sinx=\dfrac{\sqrt{2}}{8}\)
\(\Leftrightarrow sinx.cosx\left(sin^2x-cos^2x\right)=\dfrac{\sqrt{2}}{8}\)
\(\Leftrightarrow\dfrac{-1}{2}sin2x.cos2x=\dfrac{\sqrt{2}}{8}\Leftrightarrow sin2x.cos2x=-\dfrac{1}{2\sqrt{2}}\)
\(\Leftrightarrow\dfrac{1}{2}sin4x=\dfrac{-1}{2\sqrt{2}}\Leftrightarrow sin4x=\dfrac{-\sqrt{2}}{2}=sin\left(\dfrac{-\pi}{4}\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}4x=\dfrac{-\pi}{4}+k2\pi\\4x=\pi+\dfrac{\pi}{4}+k2\pi\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-\pi}{16}+\dfrac{k\pi}{2}\\x=\dfrac{5\pi}{16}+\dfrac{k\pi}{2}\end{matrix}\right.\) \(\left(k\in Z\right)\)
vậy phương trình có 2 hệ nghiệm là \(x=\dfrac{-\pi}{16}+\dfrac{k\pi}{2}\) và \(x=\dfrac{5\pi}{16}+\dfrac{k\pi}{2}\) \(\left(k\in Z\right)\)
a: \(A=sinx\cdot cosx\cdot\left(sin^4x-cos^4x\right)\)
\(=\dfrac{1}{2}\cdot sin2x\cdot\left(sin^2x-cos^2x\right)\)
\(=\dfrac{1}{2}\cdot sin2x\cdot\left(-cos2x\right)\)
\(=-\dfrac{1}{2}\cdot sin2x\cdot cos2x\)
\(=\dfrac{-1}{4}\cdot sin4x=-\dfrac{1}{4}\cdot sin\left(4\cdot\dfrac{pi}{16}\right)=-\dfrac{\sqrt{2}}{8}\)
b: \(B=\left(sin^4x+cos^4x\right)+sinx\cdot cosx\left(sin^2x-cos^2x\right)\)
\(=\left(sin^2x-cos^2x\right)^2+2\cdot\left(sinx\cdot cosx\right)^2+sinx\cdot cosx\left(sin^2x-cos^2x\right)\)
\(=\left(-cos2x\right)^2+2\cdot\left(\dfrac{1}{2}\cdot sin2x\right)^2+\dfrac{1}{2}\cdot sin2x\cdot\left(-cos2x\right)\)
\(=cos^22x+\dfrac{1}{2}\cdot sin^22x-\dfrac{1}{4}\cdot sin4x\)
\(=cos^2\left(2\cdot\dfrac{pi}{48}\right)+\dfrac{1}{2}\cdot sin^2\left(2\cdot\dfrac{pi}{48}\right)-\dfrac{1}{4}\cdot sin\left(4\cdot\dfrac{pi}{48}\right)\)
\(\simeq0.93\)