\(5sin2a-6cosa=0\)

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

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

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

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

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

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

cos\(\alpha\)=\(\dfrac{+}{-}\)\(\sqrt{1-sin\alpha^2}\)=\(\dfrac{+}{-}\dfrac{\sqrt{3}}{2}\)

Theo giả thiết ta có 3 góc: \(\alpha;\beta=\alpha+\dfrac{\pi}{3};\gamma=\alpha+\dfrac{2\pi}{3}\).
Ta có:
\(tan\alpha.tan\left(\alpha+\dfrac{\pi}{3}\right)+tan\left(\alpha+\dfrac{\pi}{3}\right).tan\left(\alpha+\dfrac{2\pi}{3}\right)+\)\(tan\left(\alpha+\dfrac{2\pi}{3}\right).tan\alpha\)
\(=tan\alpha\left[tan\left(\alpha+\dfrac{\pi}{3}\right)+tan\left(\alpha+\dfrac{2\pi}{3}\right)\right]\)\(+tan\left(a+\dfrac{\pi}{3}\right)tan\left(\alpha+\dfrac{2\pi}{3}\right)\)
\(=tan\alpha\dfrac{sin\left(2\alpha+\pi\right)}{cos\left(\alpha+\dfrac{\pi}{3}\right)cos\left(\alpha+\dfrac{2\pi}{3}\right)}\)\(+\dfrac{sin\left(\alpha+\dfrac{\pi}{3}\right)sin\left(\alpha+\dfrac{2\pi}{3}\right)}{cos\left(\alpha+\dfrac{\pi}{3}\right)cos\left(\alpha+\dfrac{2\pi}{3}\right)}\)
\(=tan\alpha\dfrac{-sin2\alpha}{cos\left(\alpha+\dfrac{\pi}{3}\right)cos\left(\alpha+\dfrac{2\pi}{3}\right)}\)\(+\dfrac{cos\dfrac{\pi}{3}-cos\left(2\alpha+\pi\right)}{2cos\left(\alpha+\dfrac{\pi}{3}\right)cos\left(\alpha+\dfrac{2\pi}{3}\right)}\)
\(=\dfrac{-2sin^2\alpha}{cos\left(\alpha+\dfrac{\pi}{3}\right)cos\left(\alpha+\dfrac{2\pi}{3}\right)}\)\(+\dfrac{\dfrac{1}{2}+cos2\alpha}{2cos\left(\alpha+\dfrac{\pi}{3}\right)cos\left(\alpha+\dfrac{2\pi}{3}\right)}\)
\(=\dfrac{\dfrac{1}{2}-4sin^2\alpha+cos2\alpha}{2cos\left(\alpha+\dfrac{\pi}{3}\right)cos\left(\alpha+\dfrac{2\pi}{3}\right)}\)
\(=\dfrac{\dfrac{1}{2}-4\left(1-cos^2\alpha\right)+2cos^2\alpha-1}{cos\dfrac{\pi}{3}+cos\left(2\alpha+\pi\right)}\)
\(=\dfrac{6cos^2\alpha-\dfrac{9}{2}}{\dfrac{1}{2}-cos2\alpha}\)
\(=\dfrac{3\left(2cos^2\alpha-\dfrac{3}{2}\right)}{\dfrac{1}{2}-\left(2cos^2\alpha-1\right)}=\dfrac{3\left(2cos^2\alpha-\dfrac{3}{2}\right)}{\dfrac{3}{2}-2cos^2\alpha}=-3\).

\(4cos\alpha.cos\beta cos\gamma=4cos\alpha cos\left(\alpha+\dfrac{\pi}{3}\right)cos\left(\alpha+\dfrac{2\pi}{3}\right)\)
\(=4cos\alpha.\dfrac{1}{2}\left(cos\dfrac{\pi}{3}+cos\left(2\alpha+\pi\right)\right)\)
\(=4cos\alpha.\dfrac{1}{2}\left(\dfrac{1}{2}-cos2\alpha\right)\)
\(=cos\alpha-2cos\alpha.cos2\alpha\)
\(=cos\alpha-\left(cos\alpha+cos3\alpha\right)\)
\(=-cos3\alpha\)
\(=cos\left(\pi+3\alpha\right)\)
\(=cos3\left(\dfrac{\pi}{3}+\alpha\right)\)
\(=cos3\beta\) (đpcm).

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:

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

\(P=\frac{sina+cosa}{sina-cosa}=\frac{\frac{sina}{sina}+\frac{cosa}{sina}}{\frac{sina}{sina}-\frac{cosa}{sina}}=\frac{1+cota}{1-cota}=\frac{1+2}{1-2}=-3\)