a: \(y=\sqrt3\cdot\sin x+cosx-2\)
\(=2\left(\frac{\sqrt3}{2}\cdot\sin x+\frac12\cdot cosx-1\right)\)
\(=2\left\lbrack\sin\left(x+\frac{\pi}{6}\right)-1\right\rbrack=2\cdot\sin\left(x+\frac{\pi}{6}\right)-2\)
Ta có: \(-1\le\sin\left(x+\frac{\pi}{6}\right)\le1\)
=>\(-2\le2\cdot\sin\left(x+\frac{\pi}{6}\right)\le2\)
=>\(-2-2\le2\cdot\sin\left(x+\frac{\pi}{6}\right)-2\le2-2\)
=>-4<=y<=0
y min =-4 khi \(\sin\left(x+\frac{\pi}{6}\right)=-1\)
=>\(x+\frac{\pi}{6}=-\frac{\pi}{2}+k2\pi\)
=>\(x=-\frac{\pi}{2}-\frac{\pi}{6}+k2\pi=-\frac23\pi+k2\pi\)
y max=0 khi \(\sin\left(x+\frac{\pi}{6}\right)=1\)
=>\(x+\frac{\pi}{6}=\frac{\pi}{2}+k2\pi\)
=>\(x=\frac{\pi}{3}+k2\pi\)
b: \(y=\sin x-cos\left(x-\frac{\pi}{6}\right)+1\)
\(=\sin x-\left\lbrack cosx\cdot cosx\left(\frac{\pi}{6}\right)+\sin x\cdot\sin\left(\frac{\pi}{6}\right)\right\rbrack+1\)
\(=\sin x-\frac{\sqrt3}{2}\cdot cosx-\frac12\cdot\sin x+1=\frac12\cdot\sin x-\frac{\sqrt3}{2}\cdot cosx+1\)
\(=\sin\left(x-\frac{\pi}{3}\right)+1\)
Ta có: \(-1\le\sin\left(x-\frac{\pi}{3}\right)\le1\)
=>\(-1+1\le\sin\left(x-\frac{\pi}{3}\right)+1\le1+1\)
=>0<=y<=2
y min =0 khi \(\sin\left(x-\frac{\pi}{3}\right)=-1\)
=>\(x-\frac{\pi}{3}=-\frac{\pi}{2}+k2\pi\)
=>\(x=-\frac{\pi}{2}+\frac{\pi}{3}+k2\pi=-\frac{\pi}{6}+k2\pi\)
y max=2 khi \(\sin\left(x-\frac{\pi}{3}\right)=1\)
=>\(x-\frac{\pi}{3}=\frac{\pi}{2}+k2\pi\)
=>\(x=\frac56\pi+k2\pi\)
c: \(y=3\cdot\sin x+4\cdot cosx-2\)
\(=5\cdot\left\lbrack\sin x\cdot\frac35+\frac45\cdot cosx\right\rbrack-2\)
\(=5\cdot\sin\left(x+\alpha\right)-2\) (với cosa=3/5; sin a=4/5)
\(-1<=\sin\left(x+a\right)\le1\)
=>\(-5\le5\cdot\sin\left(x+a\right)\le5\)
=>\(-5-2\le5\cdot\sin\left(x+a\right)-2\le5-2\)
=>-7<=y<=3
y min =-7 khi sin(x+a)=-1
=>\(x+a=-\frac{\pi}{2}+k2\pi\)
=>\(x=-a-\frac{\pi}{2}+k2\pi\)
y max=3 khi sin(x+a)=1
=>\(x+a=\frac{\pi}{2}+k2\pi\)
=>\(x=-a+\frac{\pi}{2}+k2\pi\)
