\(\sqrt{2}sinx=2sin2x.cos2x-cos2x\)

\(\Leftrightarrow\sqrt{2}sinx=sin4x-cos4x\)

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{12}+\dfrac{k\pi}{3}\\x=\dfrac{\pi}{4}+\dfrac{k2\pi}{5}\end{matrix}\right.\)

Vậy phương trình có tập nghiệm  (k ∈ Z).

`1)sin^2 x+3sin x-cos^2 x=-2`

`<=>sin^2 x+3sin x-1+sin^2 x+2=0`

`<=>2sin^2 x+3sin x+1=0`

`<=>[(sin x=-1),(sin x=-1/2):}`

`<=>[(x=-\pi/2 +k2\pi),(x=-\pi/6 +k2\pi),(x=[7\pi]/6+k2\pi):}`   `(k in ZZ)`

`2)sin^2 x+sin x-cos^2 x=0`

`<=>sin^2 x+sin x-1+sin^2 x=0`

`<=>2sin^2 x+sin x-1=0`

`<=>[(sin x=-1),(sin x=1/2):}`

`<=>[(x=-\pi/2 +k2\pi),(x=\pi/6 +k2\pi),(x=[5\pi]/6 +k2\pi):}`    `(k in ZZ)`

ĐK cos2x khác 0
1+tan^2(x) là 1/cos^2(x)

sin2x là 2sinxcosx đạp lên là ra \(2tanx=0\)

\(\cos3x+2\sin2x-\cos x=0\Leftrightarrow2\sin2x\left(1-\sin x\right)=0\)

\(\Leftrightarrow\begin{cases}\sin2x=0\\\sin x=0\end{cases}\)

\(\Leftrightarrow\begin{cases}x=k\frac{\pi}{2}\\x=\frac{\pi}{2}+k2\pi\end{cases}\)

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

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

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

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

\(\Leftrightarrow...\)