lm trên symbolab.com

\(\left(2\sin x-1\right)\left(2\sin2x+1\right)=3-4\cos^2x\)

\(\Leftrightarrow\left(2\sin x-1\right)\left(2\sin2x+1\right)=3-4\left(2-\sin^2x\right)\)

\(\Leftrightarrow\left(2\sin x-1\right)\left(2\sin2x+1\right)=4\sin^2x-1\)

\(\Leftrightarrow\left(2\sin x-1\right)\left(2\sin2x+1\right)=\left(2\sin x-1\right)\left(2\sin x+1\right)\)

\(\Leftrightarrow2\sin2x+1=2\sin x+1\)

\(\Leftrightarrow\sin2x=\sin x\)

\(\Leftrightarrow\sin2x-\sin x=0\)

\(\Leftrightarrow2\cos\frac{3}{2}-\cos\frac{x}{2}=0\)

\(\Leftrightarrow\orbr{\begin{cases}\cos\frac{3}{2}=0\\\cos\frac{x}{2}=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}\frac{3x}{2}=\frac{\pi}{2}+k2\pi\\\frac{x}{2}=\frac{\pi}{2}+k2\pi\end{cases}\left(k\inℤ\right)}\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{\pi}{3}+\frac{2\pi}{3}k\\x=\pi+4k\pi\end{cases}\left(k\inℤ\right)}\)

b) \(2sin^2x-3sinxcosx+cos^2x=0\)

\(\Leftrightarrow2tan^2x-3tanx+1=0\left(cosx\ne0\Leftrightarrow x\ne\dfrac{\pi}{2}+k\pi\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}tanx=1\\tanx=\dfrac{1}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}tanx=tan\dfrac{\pi}{4}\\tanx=\dfrac{1}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{4}+k\pi\\x=arctan\left(\dfrac{1}{2}\right)+k\pi\end{matrix}\right.\left(k\in Z\right)\)

\(\Leftrightarrow2sin^3x+1-sin^2x-1=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}sinx=0\\sinx=\dfrac{1}{2}\end{matrix}\right.\)

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

ĐK: \(x\ne k\pi\)

\(\dfrac{1+sin2x+cos2x}{1+cot^2x}=sinx.\left(sin2x+2sin^2x\right)\)

\(\Leftrightarrow\dfrac{1+sin2x+cos2x}{\dfrac{cos^2x+sin^2x}{sin^2x}}=sinx.\left(2sinx.cosx+2sin^2x\right)\)

\(\Leftrightarrow\dfrac{1+sin2x+cos2x}{\dfrac{1}{sin^2x}}=2sin^2x.\left(cosx+sinx\right)\)

\(\Leftrightarrow1+sin2x+cos2x=2cosx+2sinx\)

\(\Leftrightarrow1+2sinx.cosx+2cos^2x-1=2cosx+2sinx\)

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

\(\Leftrightarrow\left(cosx-1\right).sin\left(x+\dfrac{\pi}{4}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cosx=1\\sin\left(x+\dfrac{\pi}{4}\right)=0\end{matrix}\right.\)

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

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

Vậy phương trình có tập nghiệm 

 (k ∈ Z)

`2sin^2x+\sqrt3sin2x=3`

`<=>2. (1-cos2x)/2 + \sqrt3sin2x=3`

`<=>\sqrt3sin2x-cos2x=2`

`<=> \sqrt3/2 sin2x-1/2 cos2x=1`

`<=>sin (2x-π/6) = 1`

`<=> 2x-π/6=π/2+k2π`

`<=> x=π/3+kπ (k \in ZZ)`.

\(\Leftrightarrow1-cos2x+\sqrt{3}sin2x=3\)

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

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

\(\Leftrightarrow sin\left(2x-\dfrac{\pi}{6}\right)=1\)

\(\Leftrightarrow2x-\dfrac{\pi}{6}=\dfrac{\pi}{2}+k2\pi\)

\(\Leftrightarrow x=\dfrac{\pi}{3}+k\pi\)

a: =>sin2x+2*(1-cos2x)/2=2

=>sin2x-cos2x=1

=>căn 2*sin(2x-pi/4)=1

=>2x-pi/4=pi/4+k2pi hoặc 2x-pi/4=3/4pi+k2pi

=>x=pi/4+kpi hoặc x=pi/2+kpi

b: =>2*(1+cos2x)/2+1/2*sin2x-1/2(1-cos2x)=0

=>1+cos2x+1/2*sin2x-1/2+1/2cos2x=0

=>1/2*sin2x+3/2*cos2x=-1/2

=>sin(2x+a)=-cos(a)=cos(pi-a)

=>sin(2x+a)=sin(-pi/2+a)

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

=>x=-pi/4+kpi hoặc x=3/4pi-a+kpi