Question

Dùng công thức hạ bậc để giải các pt sau:

a)4cos²(2x - 1) = 1                                             

b) 2sin² (x + 1) = 1

c) cos² 3x + sin² 4x = 1

 d)  2cosx + 1 = 0

GIÚP MÌNH VỚI MÌNH ĐANG CẦN GẤP

 

 

Answer 1

a.

\(\Leftrightarrow2-2cos\left(4x-2\right)=1\)

\(\Leftrightarrow cos\left(4x-2\right)=\dfrac{1}{2}\)

\(\Rightarrow\left[{}\begin{matrix}4x-2=\dfrac{\pi}{3}+k2\pi\\4x-2=-\dfrac{\pi}{3}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}+\dfrac{\pi}{12}+\dfrac{k\pi}{2}\\x=\dfrac{1}{2}-\dfrac{\pi}{12}+\dfrac{k\pi}{2}\end{matrix}\right.\)

b.

\(\Leftrightarrow\dfrac{1}{2}+\dfrac{1}{2}cos6x+\dfrac{1}{2}-\dfrac{1}{2}cos8x=1\)

\(\Leftrightarrow cos6x=cos8x\)

\(\Leftrightarrow\left[{}\begin{matrix}8x=6x+k2\pi\\8x=-6x+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=k\pi\\x=\dfrac{k\pi}{7}\end{matrix}\right.\)

\(\Rightarrow x=\dfrac{k\pi}{7}\)

Answer 2

`a)4cos^2(2x-1)=1`

`<=>4[1+cos(4x-2)]/2=1`

`<=>2(1+cos(4x-2))=1`

`<=>2cos(4x-2)=-1`

`<=>cos(4x-2)=-1/2`

`<=>[(4x-2=[2\pi]/3+k2\pi),(4x-2=[-2\pi]/3+k2\pi):}`

`<=>[(x=1/2+\pi/6+k\pi/2),(x=1/2-\pi/6+k\pi/2):}`   `(k in ZZ)`

__________________________________________

`c)cos^2 3x+sin^2 4x=1`

`<=>[1+cos 6x]/2+[1-cos 8x]/2=1`

`<=>1+cos 6x+1-cos 8x=2`

`<=>cos 8x=cos 6x`

`<=>[(8x=6x+k2\pi),(8x=-6x+k2\pi):}`

`<=>[(x=k\pi),(x=k\pi/7):}`    `(k in ZZ)`