a: \(\left(\sin a+cosa\right)^2=\sin^2a+cos^2a+2\cdot\sin a\cdot cosa\)
\(=1+2\cdot\frac{12}{25}=\frac{49}{25}\)
=>\(\left[\begin{array}{l}\sin a+cosa=\frac75\\ sina+cosa=-\frac75\end{array}\right.\)
\(\sin^3a+cos^3a=\left(\sin a+cosa\right)\left(sin^2a+cos^2a-\sin a\cdot cosa\right)\)
\(=\left(\sin a+cosa\right)\left(1-\frac{12}{25}\right)=\frac{13}{25}\cdot\left(\sin a+cosa\right)\)
\(=\left[\begin{array}{l}\frac{13}{25}\cdot\frac75=\frac{91}{125}\\ \frac{13}{25}\cdot\frac{-7}{5}=\frac{-91}{125}\end{array}\right.\)
b: Đặt \(a=\sin^2x;b=\sin^2x\)
Theo đề, ta có: \(\begin{cases}a+b=1\\ 3a^2+b^2=\frac34\end{cases}\Rightarrow\begin{cases}b=1-a\\ 3a^2+b^2=\frac34\end{cases}\)
\(3a^2+b^2=\frac34\)
=>\(3a^2+\left(1-a\right)^2=\frac34\)
=>\(3a^2+a^2-2a+1-\frac34=0\)
=>\(4a^2-2a+\frac14=0\)
=>\(\left(2a-\frac12\right)^2=0\)
=>\(2a-\frac12=0\)
=>\(a=\frac14\)
=>\(b=1-\frac14=\frac34\)
\(A=\sin^4x+3\cdot cos^4x\)
\(=a^2+3b^2=\frac{1}{16}+3\cdot\left(\frac34\right)^2=\frac{1}{16}+3\cdot\frac{9}{16}=\frac{1+27}{16}=\frac{28}{16}=\frac74\)
