1: \(cos^2\left(x-\frac{\pi}{5}\right)=\sin^2\left(2x+\frac45\pi\right)\)

=>\(\left[\begin{array}{l}cos\left(x-\frac{\pi}{5}\right)=\sin\left(2x+\frac45\pi\right)=cos\left(\frac{\pi}{2}-2x-\frac45\pi\right)=cos\left(-2x-\frac{3}{10}\pi\right)\\ cos\left(x-\frac{\pi}{5}\right)=-\sin\left(2x+\frac45\pi\right)=\sin\left(-2x-\frac45\pi\right)=cos\left(\frac{\pi}{2}+2x+\frac45\pi\right)=cos\left(2x+\frac{13}{10}\pi\right)\end{array}\right.\)

TH1: \(cos\left(x-\frac{\pi}{5}\right)=cos\left(-2x-\frac{3}{10}\pi\right)\)

=>\(\left[\begin{array}{l}x-\frac{\pi}{5}=-2x-\frac{3}{10}\pi+k2\pi\\ x-\frac{\pi}{5}=2x+\frac{3}{10}\pi+k2\pi\end{array}\right.\)

=>\(\left[\begin{array}{l}3x=-\frac{3}{10}\pi+\frac{\pi}{5}+k2\pi=-\frac{1}{10}\pi+k2\pi\\ -x=\frac{3}{10}\pi+\frac{\pi}{5}+k2\pi=-\frac12\pi+k2\pi\end{array}\right.\Rightarrow\left[\begin{array}{l}x=-\frac{1}{30}\pi+\frac{k2\pi}{3}\\ x=\frac12\pi-k2\pi\end{array}\right.\)

TH2: \(cos\left(x-\frac{\pi}{5}\right)=cos\left(2x+\frac{13}{10}\pi\right)\)

=>\(\left[\begin{array}{l}2x+\frac{13}{10}\pi=x-\frac{\pi}{5}+k2\pi\\ 2x+\frac{13}{10}\pi=-x+\frac{\pi}{5}+k2\pi\end{array}\right.\Rightarrow\left[\begin{array}{l}2x-x=-\frac{\pi}{5}-\frac{13}{10}\pi+k2\pi\\ 2x+x=\frac{\pi}{5}-\frac{13}{10}\pi+k2\pi\end{array}\right.\)

=>\(\left[\begin{array}{l}x=-\frac{15}{10}\pi+k2\pi=-\frac32\pi+k2\pi\\ 3x=-\frac{11}{10}\pi+k2\pi\end{array}\right.\Rightarrow\left[\begin{array}{l}x=-\frac32\pi+k2\pi\\ x=-\frac{11}{30}\pi+\frac{k2\pi}{3}\end{array}\right.\)

2: \(\sin3x=\sqrt2\cdot cos\left(x-\frac{\pi}{5}\right)+cos3x\)

=>\(\sin3x-cos3x=\sqrt2\cdot cos\left(x-\frac{\pi}{5}\right)\)

=>\(\sqrt2\cdot\sin\left(3x-\frac{\pi}{4}\right)=\sqrt2\cdot cos\left(x-\frac{\pi}{5}\right)\)

=>\(\sin\left(3x-\frac{\pi}{4}\right)=cos\left(x-\frac{\pi}{5}\right)=\sin\left(\frac{\pi}{2}-x+\frac{\pi}{5}\right)=\sin\left(-x+\frac{7}{10}\pi\right)\)

=>\(\left[\begin{array}{l}3x-\frac{\pi}{4}=-x+\frac{7}{10}\pi+k2\pi\\ 3x-\frac{\pi}{4}=\pi+x-\frac{7}{10}\pi+k2\pi=x+\frac{3}{10}\pi+k2\pi\end{array}\right.\)

=>\(\left[\begin{array}{l}4x=\frac{7}{10}\pi+\frac{\pi}{4}+k2\pi=\frac{19}{20}\pi+k2\pi\\ 2x=\frac{3}{10}\pi+\frac{\pi}{4}+k2\pi=\frac{11}{20}\pi+k2\pi\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac{19}{80}\pi+\frac{k\pi}{2}\\ x=\frac{11}{40}\pi+k\pi\end{array}\right.\)

\(\Leftrightarrow3sinx-4sin^3x+4cos^3x-3cosx+2cosx=0\)

\(\Leftrightarrow3sinx-cosx-4sin^3x+4cos^3x=0\)

Với \(cosx=0\) ko phải nghiệm, với \(cosx\ne0\) chia 2 vế cho \(cos^3x\)

\(\Leftrightarrow3tanx\left(1+tan^2x\right)-\left(1+tan^2x\right)-4tan^3x+4=0\)

\(\Leftrightarrow-tan^3x-tan^2x+3tanx+3=0\)

\(\Leftrightarrow-tan^2x\left(tanx+1\right)+3\left(tanx+1\right)=0\)

\(\Leftrightarrow\left(tanx+1\right)\left(3-tan^2x\right)=0\)

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

Tới đây chắc bạn hoàn thành được phần còn lại

cảm ơn nha

1: Chu kì của hàm số là: \(T=\frac{2\pi}{3}\)

3: Chu kì của hàm số là: \(T=\frac{\pi}{1}=\pi\)

5: Chu kì của hàm số là \(T=\pi:\frac15=5\pi\)

a: sin 3x-cos3x+\(\sqrt3=0\)

=>\(\sin3x-cos3x=-\sqrt3\)

=>\(\sqrt2\cdot\sin\left(3x-\frac{\pi}{4}\right)=-\sqrt3\)

=>\(\sin\left(3x-\frac{\pi}{4}\right)=-\sqrt{\frac32}<-1\)

=>Phương trình không có nghiệm

b: sin x=căn 2

mà căn 2>1

nên x∈∅

=>Tập nghiệm là S=∅

c: \(\sin2x=\frac{\sqrt3}{2}\)

=>\(\left[\begin{array}{l}2x=\frac{\pi}{3}+k2\pi\\ 2x=\pi-\frac{\pi}{3}+k2\pi=\frac23\pi+k2\pi\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac{\pi}{6}+k\pi\\ x=\frac{\pi}{3}+k\pi\end{array}\right.\)

TH1: \(x=\frac{\pi}{6}+k\pi\)

\(x\in\left\lbrack-\pi;2\pi\right\rbrack\)

=>\(\frac{\pi}{6}+k\pi\in\left\lbrack-\pi;2\pi\right\rbrack\)

=>\(k+\frac16\in\left\lbrack-1;2\right\rbrack\)

=>\(k\in\left\lbrack-\frac76;\frac{11}{6}\right\rbrack\)

mà k nguyên

nên k∈{-1;0;1}

=>Có 3 nghiệm trong trường hợp này(1)

TH2: \(x=\frac{\pi}{3}+k\pi\)

x\(\in\left\lbrack-\pi;2\pi\right\rbrack\)

=>\(\frac{\pi}{3}+k\pi\in\left\lbrack-\pi;2\pi\right\rbrack\)

=>\(k+\frac13\in\left\lbrack-1;2\right\rbrack\)

=>k∈[-4/3;5/3]

mà k nguyên

nên k∈{-1;0;1}

=>Có 3 nghiệm trong trường hợp này(2)

Từ (1),(2) suy ra có 3+3=6 nghiệm

b: \(cos\left(x+\frac{\pi}{5}\right)\cdot cos\left(x-\frac{\pi}{5}\right)=cos\left(\frac{2\pi}{5}\right)\)

=>\(\frac12\cdot\left\lbrack cos\left(x+\frac{\pi}{5}+x-\frac{\pi}{5}\right)+cos\left(x+\frac{\pi}{5}-x+\frac{\pi}{5}\right)\right\rbrack=cos\left(\frac{2\pi}{5}\right)\)

=>\(\frac12\cdot\left\lbrack cos2x+cos\left(\frac{2\pi}{5}\right)\right\rbrack=cos\left(\frac{2\pi}{5}\right)\)

=>\(cos2x+cos\left(\frac{2\pi}{5}\right)=2\cdot cos\left(\frac{2\pi}{5}\right)\)

=>cos2x=\(cos\left(\frac{2\pi}{5}\right)\)

=>\(\left[\begin{array}{l}2x=\frac25\pi+k2\pi\\ 2x=-\frac25\pi+k2\pi\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac15\pi+k\pi\\ x=-\frac15\pi+k\pi\end{array}\right.\)

b.

\(\Leftrightarrow\sqrt{2}cos\left(3x+\frac{\pi}{4}\right)=-\sqrt{2}\)

\(\Leftrightarrow cos\left(3x+\frac{\pi}{4}\right)=-1\)

\(\Leftrightarrow3x+\frac{\pi}{4}=\pi+k2\pi\)

\(\Leftrightarrow x=...\)

c.

\(\Leftrightarrow\frac{\sqrt{3}}{2}sin2x+\frac{1}{2}cos2x=-\frac{\sqrt{2}}{2}\)

\(\Leftrightarrow sin\left(2x+\frac{\pi}{6}\right)=-\frac{\sqrt{2}}{2}\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+\frac{\pi}{6}=-\frac{\pi}{4}+k2\pi\\2x+\frac{\pi}{6}=\frac{5\pi}{4}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow...\)

Tham khảo

Đặt cosx-sinx=a. Thay vào giải pt ta tìm được: a=1

<=> cosx-sinx=1 

ròi bị xóa luôn

a.

\(\left(sin^2x+cos^2x\right)^3-3sin^2x.cos^2x\left(sin^2x+cos^2x\right)=cos2x+\dfrac{1}{16}\)

\(\Leftrightarrow1-\dfrac{3}{4}sin^22x=cos2x+\dfrac{1}{16}\)

\(\Leftrightarrow\dfrac{15}{16}-\dfrac{3}{4}\left(1-cos^22x\right)=cos2x\)

\(\Leftrightarrow\dfrac{3}{4}cos^22x-cos2x+\dfrac{3}{16}=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cos2x=\dfrac{4-\sqrt{7}}{6}\\cos2x=\dfrac{4+\sqrt{7}}{6}>1\left(loại\right)\end{matrix}\right.\)

\(\Rightarrow x=\pm\dfrac{1}{2}arccos\left(\dfrac{4-\sqrt{7}}{6}\right)+k\pi\)

b.

\(\left(sin^2\dfrac{x}{2}+cos^2\dfrac{x}{2}\right)^2-2sin^2\dfrac{x}{2}cos^2\dfrac{x}{2}=\dfrac{5}{2}-2sinx\)

\(\Leftrightarrow1-\dfrac{1}{2}sin^2x=\dfrac{5}{2}-2sinx\)

\(\Leftrightarrow\dfrac{1}{2}sin^2x-2sinx+\dfrac{3}{2}=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx=1\\sinx=3\left(loại\right)\end{matrix}\right.\)

\(\Leftrightarrow x=\dfrac{\pi}{2}+k2\pi\)

\(\Leftrightarrow3\sin x-4\sin^3x+4\cos^3x-3\cos x-2\cos x+2\sin x+1=0\)\(\Leftrightarrow4\left[\left(\cos x-\sin x\right)^3+3\cos x.\sin x\left(\cos x-\sin x\right)\right]-5\left(\cos x-\sin x\right)+1=0\)\(\Leftrightarrow4\left[\left(\cos x-\sin x\right)^3+3\dfrac{\left(\cos x-\sin x\right)^2-1}{2}\left(\cos x-\sin x\right)\right]-5\left(\cos x-\sin x\right)+1=0\)Đặt cosx-sinx=a. Thay vào giải pt ta tìm được: a=1

<=> cosx-sinx=1 

\(\Leftrightarrow\cos x.\sin\dfrac{\pi}{4}-\sin x.\cos\dfrac{\pi}{4}=\dfrac{1}{\sqrt{2}}\)

\(\Leftrightarrow\sin\left(\dfrac{\pi}{4}-x\right)=\sin\dfrac{\pi}{4}\)

\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{\pi}{4}-x=\dfrac{\pi}{4}-2k\pi\Rightarrow x=2k\pi\\\dfrac{\pi}{4}-x=\pi-\dfrac{\pi}{4}-2k\pi\Rightarrow x=-\dfrac{\pi}{2}+2k\pi\end{matrix}\right.\)