1: cos(2x+pi/6)=cos(pi/3-3x)

=>2x+pi/6=pi/3-3x+k2pi hoặc 2x+pi/6=3x-pi/3+k2pi

=>5x=pi/6+k2pi hoặc -x=-1/2pi+k2pi

=>x=pi/30+k2pi/5 hoặc x=pi-k2pi

2: sin(2x+pi/6)=sin(pi/3-3x)

=>2x+pi/6=pi/3-3x+k2pi hoặc 2x+pi/6=pi-pi/3+3x+k2pi

=>5x=pi/6+k2pi hoặc -x=2/3pi-pi/6+k2pi

=>x=pi/30+k2pi/5 hoặc x=-1/2pi-k2pi

1) \(cos\left(2x+\dfrac{\pi}{6}\right)=cos\left(\dfrac{\pi}{3}-2x\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+\dfrac{\pi}{6}=\dfrac{\pi}{3}-3x+k2\pi\\2x+\dfrac{\pi}{6}=-\dfrac{\pi}{3}+3x+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}5x=\dfrac{\pi}{3}-\dfrac{\pi}{6}+k2\pi\\3x-2x=\dfrac{\pi}{3}+\dfrac{\pi}{6}-k2\pi\end{matrix}\right.\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{30}+\dfrac{k2\pi}{5}\\x=\dfrac{\pi}{2}-k2\pi\end{matrix}\right.\) \(\left(k\in N\right)\)

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

Lời giải:

$\tan (\frac{\pi}{2}+x)-3\tan ^2x=\frac{\cos 2x-1}{\cos ^2x}=\frac{2\cos ^2x-2}{\cos ^2x}=\frac{2(\cos ^2x-1)}{\cos ^2x}$

$=\frac{-2\sin ^2x}{\cos ^2x}=-2\tan ^2x$

$\Leftrightarrow \tan (x+\frac{\pi}{2})=\tan ^2x$

Dễ thấy $\tan x=0$ không thỏa mãn nên $\tan x\neq 0$. Do đó pt $\Leftrightarrow \tan ^2x=\tan [\pi +(x-\frac{\pi}{2})]=\tan (x-\frac{\pi}{2})=-\tan (\frac{\pi}{2}-x)=-\cot x =\frac{-1}{\tan x}$

$\Rightarrow \tan ^3x=-1$

$\Rightarrow \tan x=-1$

$\Rightarrow x=\frac{-\pi}{4}+k\pi$ với $k$ nguyên.

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

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

Vậy...

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.

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

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

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

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

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

c.

\(\Leftrightarrow\dfrac{1}{2}cos6x+\dfrac{1}{2}cos4x=\dfrac{1}{2}cos6x+\dfrac{1}{2}cos2x+4-3\left(\dfrac{1}{2}-\dfrac{1}{2}cos2x\right)\)

\(\Leftrightarrow\dfrac{1}{2}\left(2cos^22x-1\right)=\dfrac{1}{2}cos2x+\dfrac{5}{2}+\dfrac{3}{2}cos2x\)

\(\Leftrightarrow cos^22x-2cos2x-3=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cos2x=-1\\cos2x=3\left(loại\right)\end{matrix}\right.\)

\(\Leftrightarrow2x=\pi+k2\pi\)

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

\(cos\left(2x+\dfrac{\pi}{3}\right)+cos\left(x-\dfrac{\pi}{3}\right)=0\)

\(\Leftrightarrow2cos\dfrac{3x}{2}.cos\left(\dfrac{x}{2}+\dfrac{\pi}{3}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cos\dfrac{3x}{2}=0\\cos\left(\dfrac{x}{2}+\dfrac{\pi}{3}\right)=0\end{matrix}\right.\)

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

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

a, ĐK: \(x\ne\dfrac{5\pi}{6}+k2\pi;x\ne\dfrac{\pi}{6}+k2\pi\)

\(\dfrac{2sin^2\left(\dfrac{3x}{2}-\dfrac{\pi}{4}\right)+\sqrt{3}cos^3x\left(1-3tan^2x\right)}{2sinx-1}=-1\)

\(\Leftrightarrow2sin^2\left(\dfrac{3x}{2}-\dfrac{\pi}{4}\right)+\sqrt{3}cos^3x\left(1-3tan^2x\right)=1-2sinx\)

\(\Leftrightarrow-cos\left(3x-\dfrac{\pi}{2}\right)+\sqrt{3}cos^3x.\dfrac{cos^2x-3sin^2x}{cos^2x}=-2sinx\)

\(\Leftrightarrow-sin3x+\sqrt{3}cosx.\left(cos^2x-3sin^2x\right)=-2sinx\)

\(\Leftrightarrow-sin3x+\sqrt{3}cosx.\left(4cos^2x-3\right)=-2sinx\)

\(\Leftrightarrow-sin3x+\sqrt{3}cos3x=-2sinx\)

\(\Leftrightarrow\dfrac{1}{2}sin3x-\dfrac{\sqrt{3}}{2}cos3x-sinx=0\)

\(\Leftrightarrow sin\left(3x-\dfrac{\pi}{3}\right)-sinx=0\)

\(\Leftrightarrow2cos\left(2x-\dfrac{\pi}{6}\right)sin\left(x-\dfrac{\pi}{6}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cos\left(2x-\dfrac{\pi}{6}\right)=0\\sin\left(x-\dfrac{\pi}{6}\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-\dfrac{\pi}{6}=\dfrac{\pi}{2}+k\pi\\x-\dfrac{\pi}{6}=k\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{3}+\dfrac{k\pi}{2}\\x=\dfrac{\pi}{6}+k\pi\end{matrix}\right.\)

Đối chiếu điều kiện ta được:

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{3}+k\pi\\x=\dfrac{7\pi}{6}+k2\pi\\x=-\dfrac{\pi}{6}+k2\pi\end{matrix}\right.\)

Lời giải:
ĐKXĐ:.........

PT \(\Leftrightarrow (1-\sin x).\frac{\sin ^2x}{\cos ^2x}=1+\cos x\)

\(\Rightarrow (1-\sin x)\sin ^2x=\cos ^2x(1+\cos x)\)

\(\Leftrightarrow (\sin^2x-\cos ^2x)-(\sin ^3x+\cos ^3x)=0\)

\(\Leftrightarrow (\sin x+\cos x)[(\sin x-\cos x)-(\sin ^2x-\sin x\cos x+\cos ^2x)]=0\)

\(\Leftrightarrow (\sin x+\cos x)(\sin x-\cos x-1+\sin x\cos x)=0\)

\(\Leftrightarrow (\sin x+\cos x)(\sin x-1)(\cos x+1)=0\)

Đến đây thì đơn giản rồi.