Đặt \(sinx=t\in\left[-1;1\right]\)

\(\Rightarrow y=f\left(t\right)=4t^2-3t-1\)

Xét hàm \(f\left(t\right)\) trên \(\left[-1;1\right]\)

\(-\dfrac{b}{2a}=\dfrac{3}{8}\in\left[-1;1\right]\)

\(f\left(-1\right)=6\) ; \(f\left(\dfrac{3}{8}\right)=-\dfrac{25}{16}\) ; \(f\left(1\right)=0\)

\(\Rightarrow y_{min}=-\dfrac{25}{16}\) khi \(sinx=\dfrac{3}{8}\)

\(y_{max}=6\) khi \(sinx=-1\)

\(sin\in\left[-1;1\right]\Rightarrow y=3-4sinx\in\left[-1;7\right]\)

\(\Rightarrow y_{min}=-1\Leftrightarrow sinx=1\Leftrightarrow x=\dfrac{\pi}{2}+k2\pi\)

\(y_{max}=7\Leftrightarrow sinx=-1\Leftrightarrow x=-\dfrac{\pi}{2}+k2\pi\)

\(y=\sqrt{3}cosx-sinx=2\left(\dfrac{\sqrt{3}}{2}cosx-\dfrac{1}{2}sinx\right)=2cos\left(x+\dfrac{\pi}{6}\right)\)

Vì \(cos\left(x+\dfrac{\pi}{6}\right)\in\left[-1;1\right]\Rightarrow y=\sqrt{3}cosx-sinx\in\left[-2;2\right]\)

\(\Rightarrow y_{min}=-2\Leftrightarrow cos\left(x+\dfrac{\pi}{6}\right)=-1\Leftrightarrow x+\dfrac{\pi}{6}=\pi+k2\pi\Leftrightarrow x=\dfrac{5\pi}{6}+k2\pi\)

\(y_{max}=2\Leftrightarrow cos\left(x+\dfrac{\pi}{6}\right)=1\Leftrightarrow x+\dfrac{\pi}{6}=k2\pi\Leftrightarrow x=-\dfrac{\pi}{6}+k2\pi\)

Đặt \(sinx=t\left(t\in\left[-1;1\right]\right)\)

\(y=\left|sinx+cos2x\right|=\left|2sin^2x-sinx-1\right|\)

\(\Leftrightarrow y=\left|f\left(t\right)\right|=\left|2t^2-t-1\right|\)

\(f\left(-1\right)=2\Rightarrow y=2\)

\(f\left(1\right)=0\Rightarrow y=0\)

\(f\left(\dfrac{1}{4}\right)=-\dfrac{9}{8}\Rightarrow y=\dfrac{9}{8}\)

\(\Rightarrow y_{min}=0;y_{max}=2\)

21.
a) `2sin(x-30^@)-1=0`
`<=>sin(x-30^@)=1/2`
`<=> sin(x-30^@)=sin30^@`
`<=>[(x-30^@=30^@+k360^@),(x-30^@=180^@-30^@+k360^@):}`
`<=> [(x=60^@+k360^@),(x=180^@+k360^@):}`
b) `5sin^2x+3cosx+3=0`
`<=>5(1-cos^2x)+3cosx+3=0`
`<=>-5cos^2x+3cosx+8=0`
`<=>(cosx+1)(cosx=8/5)=0`
`<=>[(cosx=-1),(cosx=8/5\ (VN)):}`
`<=>x=180^@+k360^@`
22.
`-1<=sin2x<=1`
`<=>2<=3+sin2x<=4`
`=> y_(min)=2 ; y_(max)=4`

a) Ta có : -\(\sqrt{a^2+b^2}< =asinx+bcosx< =\sqrt{a^2+b^2}\)

=> \(-\sqrt{12^2+\left(-5\right)^2}< =y< =\sqrt{12^2+\left(-5\right)^2}\)

<=> \(-\sqrt{13}< =y< =\sqrt{13}\)

Vậy min=\(-\sqrt{13}\) ,max=\(\sqrt{13}\)

b) \(-\sqrt{9+16}< =3cosx-4sinx< =\sqrt{9+16}\)

<=> -5 <=3cos x -4sinx <= 5

<=> 0<= y <= 10

Vậy min=0 max=10