Question

[Nguyễn Thiện]

Câu 1 )

a) cos (π/7 - 3x) =-√3/2

b)cos x/3 = sin(2x +30⁰)

c) tan(30⁰- 2x) =√3

d) cot(30⁰- 2x) = √3

Answer 1

[Hoàng Tử Hà]

a/ \(\Leftrightarrow\cos\left(\frac{\pi}{7}-3x\right)=\cos\left(-\frac{5}{6}\pi\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}\frac{\pi}{7}-3x=-\frac{5}{6}\pi+k2\pi\\\frac{\pi}{7}-3x=\frac{5}{6}\pi+k2\pi\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{41}{126}\pi-\frac{2}{3}k\pi\\x=-\frac{29}{42}\pi-\frac{2}{3}k\pi\end{matrix}\right.\)

b/ \(\Leftrightarrow\sin\left(90^0-\frac{x}{3}\right)=\sin\left(2x+30^0\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}90^0-\frac{x}{3}=2x+30^0+k180^0\\90^0-\frac{x}{3}=180^0-2x-30^0+k180^0\end{matrix}\right.\Leftrightarrow...\)

c/ \(DKXD:\cos\left(30^0-2x\right)\ne0\Leftrightarrow30^0-2x\ne90^0+k180^0\Leftrightarrow x\ne-30^0-k90^0\)

\(\Leftrightarrow30^0-2x=60^0+k180^0\Leftrightarrow x=-15^0-k90^0\left(tm\right)\)

d/ \(DKXD:\sin\left(30^0-2x\right)\ne0\Leftrightarrow30^0-2x\ne k180^0\Leftrightarrow x\ne15^0-k90^0\)

\(\Leftrightarrow30^0-2x=30^0+k.180^0\Leftrightarrow x=-k.90^0\left(tm\right)\)