x = kπ/2;        x = kπ/5.

Phương trình đã cho tương đương với:

1 + cos 2 x 2 + 1 + cos 4 x 2 + 1 + cos 6 x 2 + 1 + cos 8 x 2 = 2 ⇔ cos 2 x + cos 4 x + cos 6 x + cos 8 x = 0 ⇔ cos 2 x + cos 4 x + cos 6 x + cos 8 x = 0 ⇔ 2 cos 3 x cos x + 2 cos 7 x cos x = 0 ⇔ 2 cos x cos 3 x + cos 7 x = 0 ⇔ 4 cos x cos 2 x cos 5 x = 0 ⇔ cos x = 0 cos 2 x = 0 cos 5 x = 0 ⇔ x = π 2 + k π x = π 4 + k π 2 x = π 10 + k π 5

Đáp án A

Chọn A
Phương trình đã cho tương đương với: 

Phương trình đã cho tương đương với:

\(1+sinx-cos2x=0\)

\(\Leftrightarrow1+sinx-\left(1-2sin^2x\right)=0\)

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

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

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

\(sin3x-sinx+cos2x=0\)

\(\Leftrightarrow2cos2x.sinx+cos2x=0\)

\(\Leftrightarrow cos2x\left(2sinx+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cos2x=0\\sinx=-\frac{1}{2}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{4}+\frac{k\pi}{2}\\x=-\frac{\pi}{6}+k2\pi\\x=\frac{7\pi}{6}+k2\pi\end{matrix}\right.\)

`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)`

Chọn A

cos 2 x − cos 2 x = 0    ⇔ cos 2 x − ( cos 2 x − sin 2 x ) =    0 ⇔ sin 2 x = 0    ⇔ sin x = 0    ⇔ sin x   =    k π ;      k ∈ Z

Tập nghiệm của phương trình   cos 2 x   –   cos 2 x =   0 trong khoảng [0,2π) là  0 ; π

cos2x = 1- sin^x 
sin2x= 2sinxcosx 

Nhóm lại bình thường và giải thôi

a/

\(\Leftrightarrow4cos^3x-3cosx+2cos^2x-1-cosx-1=0\)

\(\Leftrightarrow2cos^3x+cos^2x-2cosx-1=0\)

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

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=k\pi\\x=\pm\frac{2\pi}{3}+k2\pi\end{matrix}\right.\)

b/

\(cos\left(8sinx\right)=1\)

\(\Leftrightarrow8sinx=k2\pi\)

\(\Leftrightarrow sinx=\frac{k\pi}{4}\)

Do \(-1\le sinx\le1\Rightarrow-1\le\frac{k\pi}{4}\le1\)

\(\Rightarrow k=\left\{-1;0;1\right\}\)

\(\Rightarrow\left[{}\begin{matrix}sinx=-\frac{\pi}{4}\\sinx=0\\sinx=\frac{\pi}{4}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\pm arcsin\left(\frac{\pi}{4}\right)+k2\pi\\x=\pi\pm arcsin\left(\frac{\pi}{4}\right)+k2\pi\\x=k\pi\end{matrix}\right.\)

c/

\(\Leftrightarrow1+2cos^2x-1+cosx=0\)

\(\Leftrightarrow2cos^2x-cosx=0\)

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

\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{2}+k\pi\\x=\pm\frac{\pi}{3}+k2\pi\end{matrix}\right.\)

d/

Đặt \(\left\{{}\begin{matrix}\left|sinx\right|=a\ge0\\cosx=b\end{matrix}\right.\) ta được hệ:

\(\left\{{}\begin{matrix}a+3b=2\\a^2+b^2=1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a=2-3b\\a^2+b^2=1\end{matrix}\right.\)

\(\Rightarrow\left(2-3b\right)^2+b^2-1=0\)

\(\Rightarrow10b^2-12b+3=0\Rightarrow\left[{}\begin{matrix}b=\frac{6+\sqrt{6}}{10}\Rightarrow a=\frac{2-3\sqrt{6}}{10}\left(l\right)\\b=\frac{6-\sqrt{6}}{10}\Rightarrow a=\frac{2+3\sqrt{6}}{10}\end{matrix}\right.\)

\(\Rightarrow cosx=\frac{6-\sqrt{6}}{10}\)

\(\Rightarrow x=\pm arccos\left(\frac{6-\sqrt{6}}{10}\right)+k2\pi\)