\(\text{1) }3sinx-4cosx=1\\ \Leftrightarrow cos^2x+\left(\frac{4cosx+1}{3}\right)^2=1\\ \Leftrightarrow cosx=\frac{-4\pm6\sqrt{6}}{25}\\ \\ \Leftrightarrow x=arccos\left(\frac{-4\pm6\sqrt{6}}{25}\right)+k2\pi\)

\(2\text{) }\sqrt{3}sinx-cosx=1\\ \Leftrightarrow\frac{\sqrt{3}}{2}sinx-\frac{1}{2}cosx=\frac{1}{2}\\ \Leftrightarrow cos\frac{\pi}{6}\cdot sinx-sin\frac{\pi}{6}\cdot cosx=\frac{1}{2}\\ \Leftrightarrow sin\left(x-\frac{\pi}{6}\right)=sin\frac{\pi}{6}\\ \Leftrightarrow\left[{}\begin{matrix}x-\frac{\pi}{6}=\frac{\pi}{6}+a2\pi\\x-\frac{\pi}{6}=\frac{5\pi}{6}+b2\pi\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{3}+a2\pi\\x=\pi+b2\pi\end{matrix}\right.\)

\(3\text{) }\sqrt{3}cosx+sinx=-2\\ \Leftrightarrow\frac{\sqrt{3}}{2}cosx+\frac{1}{2}sinx=-1\\ \Leftrightarrow sin\frac{\pi}{3}\cdot cosx+cos\frac{\pi}{3}\cdot sinx=-1\\ \Leftrightarrow sin\left(x+\frac{\pi}{3}\right)=-1=sin\frac{3\pi}{2}\\ \\ \Leftrightarrow x+\frac{\pi}{3}=\frac{3\pi}{2}+k2\pi\\ \Leftrightarrow x=\frac{7\pi}{6}+k2\pi\)

\(4\text{) }cos4x-sin4x=1\\ \Leftrightarrow cos^24x+\left(cos4x-1\right)^2=1\\ \\ \Leftrightarrow\left[{}\begin{matrix}cos4x=0\\cos4x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}4x=\frac{\pi}{2}+a\pi\\4x=b2\pi\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{8}+\frac{a\pi}{4}\\x=\frac{b\pi}{2}\end{matrix}\right.\)

\(5\text{) }\sqrt{3}cos4x+sin4x-2cos3x=0\\ \Leftrightarrow\frac{\sqrt{3}}{2}cos4x+\frac{1}{2}sin4x=cos3x\\ \Leftrightarrow cos\frac{\pi}{3}\cdot cos4x+sin\frac{\pi}{3}\cdot sin4x=cos3x\\ \Leftrightarrow cos\left(4x-\frac{\pi}{3}\right)=cos3x\\ \Leftrightarrow\left[{}\begin{matrix}4x-\frac{\pi}{3}=3x+a2\pi\\4x-\frac{\pi}{3}=-3x+b2\pi\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{3}+a2\pi\\x=\frac{\pi}{21}+\frac{b2\pi}{7}\end{matrix}\right.\\ \Leftrightarrow x=\frac{\pi}{21}+\frac{k2\pi}{7}\)

\(6\text{) }cos^2x=3sin2x+3\\ \Leftrightarrow\frac{cos2x+1}{2}=3sin2x+3\)

Giải tương tự vd 1 và 4

7) Giải tương tự vd 1 và 4

\(\Leftrightarrow\frac{\sqrt{3}}{2}sin4x-\frac{1}{2}cos4x=\frac{1}{2}sinx-\frac{\sqrt{3}}{2}cosx\)

\(\Leftrightarrow sin4x.cos\frac{\pi}{6}-cos4x.sin\frac{\pi}{6}=sinx.cos\frac{\pi}{3}-cosx.sin\frac{\pi}{3}\)

\(\Leftrightarrow sin\left(4x-\frac{\pi}{6}\right)=sin\left(x-\frac{\pi}{3}\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}4x-\frac{\pi}{6}=x-\frac{\pi}{3}+k2\pi\\4x-\frac{\pi}{6}=\pi-x+\frac{\pi}{3}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{\pi}{18}+\frac{k2\pi}{3}\\x=\frac{3\pi}{10}+\frac{k2\pi}{5}\end{matrix}\right.\)

1/ \(3-4\sin^2=4\cos^2x-1\Leftrightarrow4\left(\sin^2x+\cos^2x\right)-4=0\Leftrightarrow4.1-4=0\left(ld\right)\Rightarrow dpcm\)

2/ \(\cos^4x-\sin^4x=\left(\cos^2x+\sin^2x\right)\left(\cos^2x-\sin^2x\right)=\cos^2x-\left(1-\cos^2x\right)=2\cos^2x-1=\left(1-\sin^2x\right)-\sin^2x=1-2\sin^2x\)

3/ \(\sin^4x+\cos^4x=\left(\sin^2x+\cos^2x\right)^2-2\sin^2x.\cos^2x=1-2\sin^2x.\cos^2x\)

1.

\(2sin\left(x+10^o\right)-\sqrt{12}cos\left(x+10^o\right)=3\)

\(\Leftrightarrow\dfrac{1}{2}sin\left(x+10^o\right)-\dfrac{\sqrt{3}}{2}cos\left(x+10^o\right)=\dfrac{3}{4}\)

\(\Leftrightarrow sin\left(x+50^o\right)=\dfrac{3}{4}\)

\(\Leftrightarrow\left[{}\begin{matrix}x+50^o=arcsin\left(\dfrac{3}{4}\right)+k360^o\\x+50^o=180^o-arcsin\left(\dfrac{3}{4}\right)+k360^o\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-50^o+arcsin\left(\dfrac{3}{4}\right)+k360^o\\x=130^o-arcsin\left(\dfrac{3}{4}\right)+k360^o\end{matrix}\right.\)

2.

\(\sqrt{3}sin4x-cos4x=\sqrt{3}\)

\(\Leftrightarrow\dfrac{\sqrt{3}}{2}sin4x-\dfrac{1}{2}cos4x=\dfrac{\sqrt{3}}{2}\)

\(\Leftrightarrow sin\left(4x-\dfrac{\pi}{3}\right)=\dfrac{\sqrt{3}}{2}\)

\(\Leftrightarrow\left[{}\begin{matrix}4x-\dfrac{\pi}{3}=\dfrac{\pi}{3}+k2\pi\\4x-\dfrac{\pi}{3}=\dfrac{2\pi}{3}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2\pi}{12}+\dfrac{k\pi}{2}\\x=\dfrac{\pi}{4}+\dfrac{k\pi}{2}\end{matrix}\right.\)

3.

\(sin2x-cot\dfrac{\pi}{5}.cos2x=1\)

\(\Leftrightarrow\sqrt{1+cot\dfrac{\pi}{5}}\left(\dfrac{1}{\sqrt{1+cot\dfrac{\pi}{5}}}sin2x-\dfrac{cot\dfrac{\pi}{5}}{\sqrt{1+cot\dfrac{\pi}{5}}}.cos2x\right)=1\)

\(\Leftrightarrow sin\left[2x-arccos\left(\dfrac{1}{\sqrt{1+cot\dfrac{\pi}{5}}}\right)\right]=\dfrac{1}{\sqrt{1+cot\dfrac{\pi}{5}}}\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-arccos\left(\dfrac{1}{\sqrt{1+cot\dfrac{\pi}{5}}}\right)=arcsin\left(\dfrac{1}{\sqrt{1+cot\dfrac{\pi}{5}}}\right)+k2\pi\\2x-arccos\left(\dfrac{1}{\sqrt{1+cot\dfrac{\pi}{5}}}\right)=\pi-arcsin\left(\dfrac{1}{\sqrt{1+cot\dfrac{\pi}{5}}}\right)+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}arccos\left(\dfrac{1}{\sqrt{1+cot\dfrac{\pi}{5}}}\right)+\dfrac{1}{2}arcsin\left(\dfrac{1}{\sqrt{1+cot\dfrac{\pi}{5}}}\right)+k\pi\\x=\dfrac{\pi}{2}+\dfrac{1}{2}arccos\left(\dfrac{1}{\sqrt{1+cot\dfrac{\pi}{5}}}\right)-\dfrac{1}{2}arcsin\left(\dfrac{1}{\sqrt{1+cot\dfrac{\pi}{5}}}\right)+k\pi\end{matrix}\right.\)

d: cos^2x=1

=>sin^2x=0

=>sin x=0

=>x=kpi

a: =>sin 4x=cos(x+pi/6)

=>sin 4x=sin(pi/2-x-pi/6)

=>sin 4x=sin(pi/3-x)

=>4x=pi/3-x+k2pi hoặc 4x=2/3pi+x+k2pi

=>x=pi/15+k2pi/5 hoặc x=2/9pi+k2pi/3

b: =>x+pi/3=pi/6+k2pi hoặc x+pi/3=-pi/6+k2pi

=>x=-pi/2+k2pi hoặc x=-pi/6+k2pi

c: =>4x=5/12pi+k2pi hoặc 4x=-5/12pi+k2pi

=>x=5/48pi+kpi/2 hoặc x=-5/48pi+kpi/2

Chọn A.

Ta có:

+ sin⁴x + cos⁴x = (sin²x + cos²x)² - 2sin²x.cos²x = 1 - 2sin²x.cos²x.

+ sin⁴x + cos⁴x = 1 - 3sin²x.cos²x.

Do đó

A = 3(1 - 2sin²x^.cos²x) - 2(1 - 3sin²x.cos²x) = 1.

Chứng minh các biểu thức đã cho không phụ thuộc vào x.

f(x) = 1 ⇒ f′(x) = 0