Question

1) 2sin(x+10\(^o\)) - \(\sqrt{12}\)cos(x+10\(^o\))=3
2) \(\sqrt{3}\)sin4x - cos4x =\(\sqrt{3}\)
3) sin2x - cot \(\dfrac{pi}{5}\).cos2x=1
4) cos x -\(\sqrt{3}\) sinx = -2cos3x

Answer 1

1.

\(2sin\left(x+10^o\right)-\sqrt{12}cos\left(x+10^o\right)=3\)

\(\Leftrightarrow\dfrac{1}{2}sin\left(x+10^o\right)-\dfrac{\sqrt{3}}{2}cos\left(x+10^o\right)=\dfrac{3}{4}\)

\(\Leftrightarrow sin\left(x+50^o\right)=\dfrac{3}{4}\)

\(\Leftrightarrow\left[{}\begin{matrix}x+50^o=arcsin\left(\dfrac{3}{4}\right)+k360^o\\x+50^o=180^o-arcsin\left(\dfrac{3}{4}\right)+k360^o\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-50^o+arcsin\left(\dfrac{3}{4}\right)+k360^o\\x=130^o-arcsin\left(\dfrac{3}{4}\right)+k360^o\end{matrix}\right.\)

Answer 2

2.

\(\sqrt{3}sin4x-cos4x=\sqrt{3}\)

\(\Leftrightarrow\dfrac{\sqrt{3}}{2}sin4x-\dfrac{1}{2}cos4x=\dfrac{\sqrt{3}}{2}\)

\(\Leftrightarrow sin\left(4x-\dfrac{\pi}{3}\right)=\dfrac{\sqrt{3}}{2}\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2\pi}{12}+\dfrac{k\pi}{2}\\x=\dfrac{\pi}{4}+\dfrac{k\pi}{2}\end{matrix}\right.\)

Answer 3

3.

\(sin2x-cot\dfrac{\pi}{5}.cos2x=1\)

\(\Leftrightarrow\sqrt{1+cot\dfrac{\pi}{5}}\left(\dfrac{1}{\sqrt{1+cot\dfrac{\pi}{5}}}sin2x-\dfrac{cot\dfrac{\pi}{5}}{\sqrt{1+cot\dfrac{\pi}{5}}}.cos2x\right)=1\)

\(\Leftrightarrow sin\left[2x-arccos\left(\dfrac{1}{\sqrt{1+cot\dfrac{\pi}{5}}}\right)\right]=\dfrac{1}{\sqrt{1+cot\dfrac{\pi}{5}}}\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-arccos\left(\dfrac{1}{\sqrt{1+cot\dfrac{\pi}{5}}}\right)=arcsin\left(\dfrac{1}{\sqrt{1+cot\dfrac{\pi}{5}}}\right)+k2\pi\\2x-arccos\left(\dfrac{1}{\sqrt{1+cot\dfrac{\pi}{5}}}\right)=\pi-arcsin\left(\dfrac{1}{\sqrt{1+cot\dfrac{\pi}{5}}}\right)+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}arccos\left(\dfrac{1}{\sqrt{1+cot\dfrac{\pi}{5}}}\right)+\dfrac{1}{2}arcsin\left(\dfrac{1}{\sqrt{1+cot\dfrac{\pi}{5}}}\right)+k\pi\\x=\dfrac{\pi}{2}+\dfrac{1}{2}arccos\left(\dfrac{1}{\sqrt{1+cot\dfrac{\pi}{5}}}\right)-\dfrac{1}{2}arcsin\left(\dfrac{1}{\sqrt{1+cot\dfrac{\pi}{5}}}\right)+k\pi\end{matrix}\right.\)

Answer 4

4.

\(cosx-\sqrt{3}sinx=-2cos3x\)

\(\Leftrightarrow\dfrac{1}{2}cosx-\dfrac{\sqrt{3}}{2}sinx+cos3x=0\)

\(\Leftrightarrow cos\left(x+\dfrac{\pi}{3}\right)+cos3x=0\)

\(\Leftrightarrow2cos\left(2x+\dfrac{\pi}{6}\right).cos\left(x-\dfrac{\pi}{6}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cos\left(2x+\dfrac{\pi}{6}\right)=0\\cos\left(x-\dfrac{\pi}{6}\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+\dfrac{\pi}{6}=\dfrac{\pi}{2}+k\pi\\x-\dfrac{\pi}{6}=\dfrac{\pi}{2}+k\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{6}+\dfrac{k\pi}{2}\\x=\dfrac{2\pi}{3}+k\pi\end{matrix}\right.\)