Ta có \(-1\le\sin2x\le1\)

\(\Leftrightarrow1\le-\sin2x\le-1\\ \Leftrightarrow0\le1-\sin2x\le2\\ \Leftrightarrow0\le y\le2\)

\(\Leftrightarrow y_{max}=2\\ y_{min}=0\)

24.

\(cos\left(x-\dfrac{\pi}{2}\right)\le1\Rightarrow y\le3.1+1=4\)

\(y_{max}=4\)

26.

\(y=\sqrt{2}cos\left(2x-\dfrac{\pi}{4}\right)\)

Do \(cos\left(2x-\dfrac{\pi}{4}\right)\le1\Rightarrow y\le\sqrt{2}\)

\(y_{max}=\sqrt{2}\)

b.

\(\dfrac{1}{2}sinx+\dfrac{\sqrt{3}}{2}cosx=\dfrac{1}{2}\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{\pi}{6}=\dfrac{\pi}{3}+k2\pi\\x-\dfrac{\pi}{6}=-\dfrac{\pi}{3}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{2}+k2\pi\\x=-\dfrac{\pi}{6}+k2\pi\end{matrix}\right.\)

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`

\(y=2\left(\dfrac{1}{2}sin2x+\dfrac{\sqrt{3}}{2}cos2x\right)=2sin\left(2x+\dfrac{\pi}{3}\right)\)

\(-1\le sin\left(2x+\dfrac{\pi}{3}\right)\le1\Rightarrow-2\le y\le2\)

\(y_{min}=-2\) khi \(sin\left(2x+\dfrac{\pi}{3}\right)=-1\Rightarrow x=-\dfrac{5\pi}{12}+k\pi\)

\(y_{max}=2\) khi \(sin\left(2x+\dfrac{\pi}{3}\right)=1\Rightarrow x=\dfrac{\pi}{12}+k\pi\)

1: \(y=\sqrt{3}\cdot sin^2x-\left(1-sin^2x\right)+5\)

\(=sin^2x\left(\sqrt{3}+1\right)-1+5=sin^2x\left(\sqrt{3}+1\right)+4\)

\(0< =sin^2x< =1\)

=>\(0< =sin^2x\left(\sqrt{3}+1\right)< =\sqrt{3}+1\)

=>4<=y<=căn 3+5

y min=4 khi sin^2x=0

=>sin x=0

=>x=kpi

\(y_{max}=5+\sqrt{3}\) khi \(sin^2x=1\)

=>\(cos^2x=0\)

=>cosx=0

=>\(x=\dfrac{pi}{2}+kpi\)

2: \(y=5\left[\dfrac{3}{5}sinx+\dfrac{4}{5}cosx\right]+7\)

\(=5\cdot\left[sinx\cdot cosa+cosx\cdot sina\right]+7\)(Với cosa=3/5; sin a=4/5)

\(=5\cdot sin\left(x+a\right)+7\)

-1<=sin(x+a)<=1

=>-5<=5sin(x+a)<=5

=>-5+7<=y<=5+7

=>2<=y<=12

\(y_{min}=2\) khi sin (x+a)=-1

=>x+a=-pi/2+kp2i

=>\(x=-\dfrac{pi}{2}+k2pi-a\)

\(y_{max}=12\) khi sin(x+a)=1

=>x+a=pi/2+k2pi

=>\(x=\dfrac{pi}{2}+k2pi-a\)

giúp mk vs

a.

\(y=1-cos^2x-6cosx+1=-cos^2x-6cosx+2\)

Đặt \(cosx=t\in\left[-1;1\right]\)

\(y=f\left(t\right)=-t^2-6t+2\)

Xét hàm \(f\left(t\right)=-t^2-6t+2\) trên \(\left[-1;1\right]\)

\(-\dfrac{b}{2a}=-3\notin\left[-1;1\right]\) ; \(f\left(-1\right)=7\) ; \(f\left(1\right)=-5\)

\(\Rightarrow y_{min}=-5\) khi \(cosx=1\Rightarrow x=k2\pi\)

b.

Đề là \(sin^4x+cos^4x\) hay \(sinx^4+cosx^4\) ?

\(y=sin^2x-6cosx+1=1-cos^2x-6cosx+1=-cos^2x-6cosx+2\)

\(y=-cos^2x-6cosx+7-5=\left(1-cosx\right)\left(cosx+7\right)-5\)

Do \(-1\le cosx\le1\Rightarrow\left\{{}\begin{matrix}1-cosx\ge0\\cosx+7>0\end{matrix}\right.\)

\(\Rightarrow\left(1-cosx\right)\left(cosx+7\right)\ge0\)

\(\Rightarrow y\ge0-5=-5\)

Lần sau em lưu ý khi viết biểu thức với sinx, cosx, \(\left(sinx\right)^2\) hay \(sin^2x\) nếu viết là \(sinx^2\) là sai hoàn toàn, người ta sẽ hiểu đó là \(sin\left(x^2\right)\)