Question

g/sinx + cosx . tan(pi/12)=1

h/cosx+1= sinx . tan(-pi/5)

Answer 1

g, \(sinx+cosx.tan\dfrac{\pi}{12}=1\)

\(\Leftrightarrow\sqrt{1^2+tan^2\dfrac{\pi}{12}}\left(\dfrac{1}{\sqrt{1^2+tan^2\dfrac{\pi}{12}}}sinx+\dfrac{tan\dfrac{\pi}{12}}{\sqrt{1^2+tan^2\dfrac{\pi}{12}}}cosx\right)=1\)

\(\Leftrightarrow sin\left(x+arccos\dfrac{1}{\sqrt{1^2+tan^2\dfrac{\pi}{12}}}\right)=\dfrac{1}{\sqrt{1^2+tan^2\dfrac{\pi}{12}}}\)

\(\Leftrightarrow\left[{}\begin{matrix}x+arccos\dfrac{1}{\sqrt{1^2+tan^2\dfrac{\pi}{12}}}=arcsin\dfrac{1}{\sqrt{1^2+tan^2\dfrac{\pi}{12}}}+k2\pi\\x+arccos\dfrac{1}{\sqrt{1^2+tan^2\dfrac{\pi}{12}}}=\pi-arcsin\dfrac{1}{\sqrt{1^2+tan^2\dfrac{\pi}{12}}}+k2\pi\end{matrix}\right.\)