mai đăng lại bài này nhé t làm cho h đi ngủ

\(\left(2cosx+\sqrt{3}\right)\left(cos2x+2sinx-\sqrt{3}\right)=1-4\left(1-cos^2x\right)\)

\(\Leftrightarrow\left(2cosx+\sqrt{3}\right)\left(cos2x+2sinx-\sqrt{3}\right)=4cos^2x-3\)

\(\Leftrightarrow\left(2cosx+\sqrt{3}\right)\left(cos2x+2sinx-\sqrt{3}\right)=\left(2cosx+\sqrt{3}\right)\left(2cosx-\sqrt{3}\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}cosx=-\dfrac{\sqrt{3}}{2}\Rightarrow x=...\\cos2x+2sinx-\sqrt{3}=2cosx-\sqrt{3}\left(1\right)\end{matrix}\right.\)

\(\left(1\right)\Leftrightarrow cos^2x-sin^2x-2\left(cosx-sinx\right)=0\)

\(\Leftrightarrow\left(cosx-sinx\right)\left(cosx+sinx\right)-2\left(cosx-sinx\right)=0\)

\(\Leftrightarrow\left(cosx-sinx\right)\left(cosx+sinx-2\right)=0\)

\(\Leftrightarrow...\)

ĐKXĐ: \(x\ne\dfrac{k\pi}{2}\)

\(\dfrac{cosx}{sinx}-1=\dfrac{cos^2x-sin^2x}{1+\dfrac{sinx}{cosx}}+sin^2x-sinx.cosx\)

\(\Leftrightarrow\dfrac{cosx-sinx}{sinx}=cosx\left(cosx-sinx\right)-sinx\left(cosx-sinx\right)\)

\(\Leftrightarrow\left(cosx-sinx\right)\left(\dfrac{1}{sinx}-cosx+sinx\right)=0\)

\(\Leftrightarrow\left(cosx-sinx\right)\left(1-sinx.cosx+sin^2x\right)=0\)

\(\Leftrightarrow\left(cosx-sinx\right)\left(3-sin2x-cos2x\right)=0\)

\(\Leftrightarrow\left(cosx-sinx\right)\left(3-\sqrt{2}sin\left(2x+\dfrac{\pi}{4}\right)\right)=0\)

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.

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

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

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

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

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

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

1)pt\(\Leftrightarrow sin^8x\left(1-2sin^2x\right)=cos^8x\left(2cos^2x-1\right)+\frac{5}{4}cos2x\)

\(\Leftrightarrow sin^8x.cos2x=cos^8x.cos2x+\frac{5}{4}cos2x\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}cos2x=0\Leftrightarrow x=\frac{\pi}{4}+\frac{k\pi}{2}\\sin^8x-cos^8x=\frac{5}{4}\left(\cdot\right)\end{array}\right.\)

Xét (*):VT(*)\(\le sin^8x\le1\)\(\Rightarrow\)pt(*) vô ngiệm

Vậy pt có 1 họ nghiệm là \(x=\frac{\pi}{4}+\frac{k\pi}{2},k\in Z\)

2)+)sinx=0 không là nghiệm của pt

+)sinx\(\ne0\):

pt\(\Leftrightarrow16sinx.cosx.cos2x.cos4x.cos8x=1\)

\(\Leftrightarrow8sin2x.cos2x.cos4x.cos8x=1\)

\(\Leftrightarrow4sin4x.cos4x.cos8x=1\)\(\Leftrightarrow2sin8x.cos8x=1\Leftrightarrow sin16x=1\Leftrightarrow x=\frac{\pi}{32}+\frac{k\pi}{8},k\in Z\)

KL:...

a.

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

\(\Leftrightarrow2-\left(2sin\dfrac{x}{2}cos\dfrac{x}{2}\right)^2=1\)

\(\Leftrightarrow1-sin^2x=0\)

\(\Leftrightarrow cos^2x=0\)

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

b.

\(\left(sin^2x+cos^2x\right)^3-3sin^2x.cos^2x\left(sin^2x+cos^2x\right)=\dfrac{7}{16}\)

\(\Leftrightarrow1-\dfrac{3}{4}\left(2sinx.cosx\right)^2=\dfrac{7}{16}\)

\(\Leftrightarrow16-12.sin^22x=7\)

\(\Leftrightarrow3-4sin^22x=0\)

\(\Leftrightarrow3-2\left(1-cos4x\right)=0\)

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

\(\Leftrightarrow4x=\pm\dfrac{2\pi}{3}+k2\pi\)

\(\Leftrightarrow x=\pm\dfrac{\pi}{6}+\dfrac{k\pi}{2}\)

c.

\(\left(sin^2x+cos^2x\right)^3-3sin^2x.cos^2x\left(sin^2x+cos^2x\right)=cos^22x+\dfrac{1}{4}\)

\(\Leftrightarrow1-\dfrac{3}{4}\left(2sinx.cosx\right)^2=cos^22x+\dfrac{1}{4}\)

\(\Leftrightarrow3-3sin^22x=4cos^22x\)

\(\Leftrightarrow3=3\left(sin^22x+cos^22x\right)+cos^22x\)

\(\Leftrightarrow3=3+cos^22x\)

\(\Leftrightarrow cos2x=0\)

\(\Leftrightarrow x=\dfrac{\pi}{4}+\dfrac{k\pi}{2}\)

1.

\(\Leftrightarrow sin^2x\left(sinx+1\right)-2\left(1-cosx\right)=0\)

\(\Leftrightarrow\left(1-cos^2x\right)\left(sinx+1\right)-2\left(1-cosx\right)=0\)

\(\Leftrightarrow\left(1-cosx\right)\left(1+cosx\right)\left(sinx+1\right)-2\left(1-cosx\right)=0\)

\(\Leftrightarrow\left(1-cosx\right)\left(sinx+cosx+sinx.cosx-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cosx=1\Leftrightarrow...\\sinx+cosx+sinx.cosx-1=0\left(1\right)\end{matrix}\right.\)

Xét (1):

Đặt \(sinx+cosx=t\Rightarrow\left[{}\begin{matrix}\left|t\right|\le\sqrt{2}\\sinx.cosx=\frac{t^2-1}{2}\end{matrix}\right.\)

\(\Leftrightarrow t+\frac{t^2-1}{2}-1=0\)

\(\Leftrightarrow t^2+2t-3=0\Rightarrow\left[{}\begin{matrix}t=1\\t=-3\left(l\right)\end{matrix}\right.\)

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

\(\Leftrightarrow...\)

2.

\(\Leftrightarrow\sqrt{3}sinx.cosx+\sqrt{2}cos^2x+\sqrt{6}cosx=0\)

\(\Leftrightarrow cosx\left(\sqrt{3}sinx+\sqrt{2}cosx+\sqrt{6}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cosx=0\Leftrightarrow...\\\sqrt{3}sinx+\sqrt{2}cosx=-\sqrt{6}\left(1\right)\end{matrix}\right.\)

Xét (1):

Do \(\sqrt{3}^2+\sqrt{2}^2< \left(-\sqrt{6}\right)^2\) nên (1) vô nghiệm

3.

\(\Leftrightarrow4sinx.cosx-\left(1-2sin^2x\right)=7sinx+2cosx-4\)

\(\Leftrightarrow2cosx\left(2sinx-1\right)+2sin^2x-7sinx+3=0\)

\(\Leftrightarrow2cosx\left(2sinx-1\right)+\left(sinx-3\right)\left(2sinx-1\right)=0\)

\(\Leftrightarrow\left(2cosx+sinx-3\right)\left(2sinx-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx=\frac{1}{2}\Leftrightarrow...\\2cosx+sinx=3\left(1\right)\end{matrix}\right.\)

Xét (1), do \(2^2+1^2< 3^2\) nên (1) vô nghiệm