Question

Answer 1

1, tan⁴x - 3tan³x + 3tan²x - 3tanx + 2 = 0

⇔ (tanx - 1)(tanx - 2)(tan²x + 1) = 0

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

2, Ta có công thức nhân ba:

sin3x = sin(2x + x) = sin2x cosx + cos2x . sinx

= 2sinx.cos²x + (1 - 2sin²x) . sinx

= 2sinx (1 - sin²x) + sinx - 2sin³x

= 2sinx - 2sin³x + sinx - 2sin³x

= 3sinx - 4sin³x

Vậy ta có phương trình

4 . [3sinx - 4sin³x - (1 - 2sin²x)] = 5sinx - 5

⇔ 4 .(3sinx - 4sin³x - 1 + 2sin²x) - 5sinx + 5 = 0

⇔ 12sinx - 16sin³x - 4 + 8sin²x - 5sinx + 5 = 0

⇔ -16sin³x + 8sin²x + 7sinx + 1 = 0

⇔\(\left[{}\begin{matrix}sinx=1\\sinx=-\dfrac{1}{4}\end{matrix}\right.\)

⇔ \(\left[{}\begin{matrix}x=\dfrac{\pi}{2}+k2\pi\\x=arcsin\left(-\dfrac{1}{4}\right)+k2\pi\\x=\pi-arcsin\left(-\dfrac{1}{4}\right)+k2\pi\end{matrix}\right.\), k ∈ Z

 

Answer 2

1.

\(tan^4x-3tan^3x+3tan^2x-3tanx+2=0\)

\(\Leftrightarrow\left(tan^3x-2tan^2x+tanx-2\right)\left(tanx-1\right)=0\)

\(\Leftrightarrow\left(tan^2x+1\right)\left(tanx-2\right)\left(tanx-1\right)=0\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=arctan2+k\pi\\x=\dfrac{\pi}{4}+k\pi\end{matrix}\right.\)