\(5sin2a-6cosa=0\)

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

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

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

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

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

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

bạn viết ddeefc ó sai ddaau k vậy sao cái cuối là +2

Ta có: \(sin^2\alpha+cos^2\alpha=1\Leftrightarrow cos^2\alpha=1-sin^2\alpha=1-\left(\dfrac{1}{2}\right)^2=\dfrac{3}{4}\)

\(P=\sqrt{2}\left(1+cot\alpha\right)\cdot cos\left(\dfrac{\pi}{4}+\alpha\right)\)

    \(=\sqrt{2}\left(1+\dfrac{cos\alpha}{sin\alpha}\right)\left(cos\dfrac{\pi}{4}\cdot cos\alpha-sin\dfrac{\pi}{4}\cdot sin\alpha\right)\)

    \(=\sqrt{2}\left(1+\dfrac{cos\alpha}{sin\alpha}\right)\left(\dfrac{\sqrt{2}}{2}\cdot cos\alpha-\dfrac{\sqrt{2}}{2}\cdot sin\alpha\right)\)

    \(=\left(1+\dfrac{cos\alpha}{sin\alpha}\right)\left(\dfrac{\sqrt{2}\cdot\sqrt{2}}{2}\cdot cos\alpha-\dfrac{\sqrt{2}\cdot\sqrt{2}}{2}\cdot sin\alpha\right)\)

    \(=\left(1+\dfrac{cos\alpha}{sin\alpha}\right)\left(cos\alpha-sin\alpha\right)\)

    \(=cos\alpha-sin\alpha+cos\alpha\cdot\dfrac{cos\alpha}{sin\alpha}-sin\alpha\cdot\dfrac{cos\alpha}{sin\alpha}\)

    \(=cos\alpha-sin\alpha+\dfrac{cos^2\alpha}{sin\alpha}-cos\alpha\)

    \(=\dfrac{cos^2\alpha}{sin\alpha}-sin\alpha\)

Thay \(cos^2\alpha=\dfrac{3}{4}\) và \(sin\alpha=\dfrac{1}{2}\) ta có:

\(P=\dfrac{3}{4}\div\dfrac{1}{2}-\dfrac{1}{2}=1\)

Ta có : \(A=\frac{\tan\alpha}{1+\tan^2\alpha}=\tan\alpha.\cos^2\alpha=\sin\alpha.\cos\alpha=\frac{3}{5}\cos\alpha\left(1\right)\)

\(\cos^2\alpha=1-\sin^2\alpha=1-\left(\frac{3}{5}\right)^2=\frac{16}{25}\)  (2)

Vì \(\alpha\in\left(\frac{\pi}{2};\pi\right)\) nên \(\cos\alpha<0\)

Do đó, từ (2) suy ra \(\cos\alpha=-\frac{4}{5}\) (3)

Thế (3) vào (1) ta được \(A=-\frac{12}{25}\)