\(sin^23x+cos^22x=1\)

\(\Leftrightarrow cos^22x=1-sin^23x=cos^23x\)

\(\Leftrightarrow\orbr{\begin{cases}cos2x=cos3x\\cos2x=-cos3x\end{cases}}\)

- \(cos2x=cos3x\)

\(\Leftrightarrow2x=\pm3x+k2\pi\left(k\inℤ\right)\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{k2\pi}{5}\left(k\inℤ\right)\\x=k2\pi\left(k\inℤ\right)\end{cases}}\)

- \(cos2x=-cos3x=cos\left(\pi-3x\right)\)

\(\Leftrightarrow2x=\pm\left(\pi-3x\right)+k2\pi\left(k\inℤ\right)\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{\pi}{5}+\frac{k2\pi}{5}\left(k\inℤ\right)\\x=\pi+k2\pi\left(k\inℤ\right)\end{cases}}\)

\(SA=SB=SC\)nên hình chiếu vuông góc từ \(S\)xuống mặt phẳng đáy là trọng tâm của tam giác \(ABC\).

Gọi \(G\)là trọng tâm tam giác \(ABC\).

\(\widehat{\left(SA,\left(ABC\right)\right)}=\widehat{SAG}\)

\(AG=cos60^o.SA=\frac{a}{2}\), \(SG=sin60^o.SA=\frac{a\sqrt{3}}{2}\)

\(\Rightarrow AB=\frac{a\sqrt{3}}{2}\)

\(S_{ABC}=\frac{\left(\frac{a\sqrt{3}}{2}\right)^2\sqrt{3}}{4}=\frac{3a^2\sqrt{3}}{4}\)

\(V_{S.ABC}=\frac{1}{3}SG.S_{ABC}=\frac{1}{3}.\frac{a\sqrt{3}}{2}.\frac{3a^2\sqrt{3}}{4}=\frac{3a^3}{8}\)

2sin^2x+5cosx+1=0

\(2\cdot\left(1-cos^2x\right)+5cosx+1=0\)   

\(-2cos^2x+5cosx+3=0\)   

\(\orbr{\begin{cases}cosx=3\left(l\right)\\cosx=-\frac{1}{2}\end{cases}}\)   

\(cosx=cos\frac{2pi}{3}\)   

\(\orbr{\begin{cases}x=\frac{2pi}{3}+k2pi\\x=\frac{-2pi}{3}+k2pi\end{cases}}\)

cos2x.tan6x=sin10x 

ĐK : cos6x khác 0 

cos2x.sin6x/cos6x=sin10x

sin6xcos2x=sin10xcos6x 

1/2(sin8x+sin4x)=1/2(sin16x+sin4x)

sin8x+sin4x=sin16x+sin4x

sin8x=sin16x

sin16x=sin8x

\(\orbr{\begin{cases}16x=8x+k2pi\\16x=\frac{pi}{2}-8x+k2pi\end{cases}}\)   

\(\orbr{\begin{cases}16x-8x=k2pi\\16x+8x=\frac{pi}{2}+k2pi\end{cases}}\)   

\(\orbr{\begin{cases}8x=k2pi\\24x=\frac{pi}{2}+k2pi\end{cases}}\)   

\(\orbr{\begin{cases}x=\frac{kpi}{4}\\x=\frac{pi}{48}+\frac{kpi}{12}\end{cases}\left(k\in Z\right)}\)

sin^2(4x)+cos^2(6x)=1

\(\frac{1-cos8x}{2}+\frac{1+cos12x}{2}=1\)   

\(\frac{1}{2}-\frac{1}{2}cos8x+\frac{1}{2}+\frac{1}{2}cos12x=1\)   

\(\frac{1}{2}cos12x-\frac{1}{2}cos8x=0\)   

\(cos12x-cos8x=0\)   

\(-2sin10xsin2x=0\)   

\(\orbr{\begin{cases}sin10x=0\\sin2x=0\end{cases}}\)   

\(\orbr{\begin{cases}10x=kpi\\2x=kpi\end{cases}}\)   

\(\orbr{\begin{cases}x=\frac{kpi}{10}\\x=\frac{kpi}{2}\end{cases}}\)   

\(x=\frac{kpi}{10}\left(k\in Z\right)\)