Question

Cho \(5\sin2\alpha-6\cos\alpha=0\) và \(0< \alpha< \dfrac{\pi}{2}\)

Tính A = \(\cos(\dfrac{\pi}{2}-\alpha)+\sin\left(2017\pi-\alpha\right)-\cot(2018\pi+\alpha)\)

Answer 1

 

\(5sin2a-6cosa=0\)

\(\Leftrightarrow sin2a=\dfrac{6}{5}cosa\)

\(\Leftrightarrow2\cdot sina\cdot cosa=\dfrac{6}{5}\cdot cosa\)

\(\Leftrightarrow cosa\left(2sina-\dfrac{6}{5}\right)=0\)

=>cosa=0 hoặc sina=3/5

hay \(a=\dfrac{\Pi}{2}+k\Pi\) hoặc \(\left[{}\begin{matrix}a=arcsin\left(\dfrac{3}{5}\right)+k2\Pi\\a=\Pi-arcsin\left(\dfrac{3}{5}\right)+k2\Pi\end{matrix}\right.\)

mà 0<a<pi/2

nên \(a=arcsin\left(\dfrac{3}{5}\right)\)

\(A=sina+sina+cota=2\cdot sina+cota\)

\(=\dfrac{38}{15}\)