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

Có \(-1\le sin\left(x+\dfrac{\pi}{6}\right)\le1\) \(\Leftrightarrow-2\le2sin\left(x+\dfrac{\pi}{6}\right)\le2\)

\(\Leftrightarrow-2\le y\le2\)

miny=-2 \(\Leftrightarrow sin\left(x+\dfrac{\pi}{6}\right)=-1\)  \(\Leftrightarrow x+\dfrac{\pi}{6}=-\dfrac{\pi}{2}+2k\pi\left(k\in Z\right)\) \(\Leftrightarrow x=-\dfrac{2\pi}{3}+k2\pi\left(k\in Z\right)\)

maxy=2\(\Leftrightarrow sin\left(x+\dfrac{\pi}{6}\right)=1\) \(\Leftrightarrow x+\dfrac{\pi}{6}=\dfrac{\pi}{2}+k2\pi\left(k\in Z\right)\)\(\Leftrightarrow x=\dfrac{\pi}{3}+k2\pi\left(k\in Z\right)\)

b) \(y=sin2x-cos2x=\sqrt{2}sin\left(2x-\dfrac{\pi}{4}\right)\)

Có \(\sqrt{2}\ge\sqrt{2}sin\left(2x-\dfrac{\pi}{4}\right)\ge-\sqrt{2}\)

\(\Leftrightarrow\sqrt{2}\ge y\ge-\sqrt{2}\)

miny=\(-\sqrt{2}\) \(\Leftrightarrow sin\left(2x-\dfrac{\pi}{4}\right)=-1\)\(\Leftrightarrow2x-\dfrac{\pi}{4}=-\dfrac{\pi}{2}+k2\pi\left(k\in Z\right)\)\(\Leftrightarrow x=-\dfrac{\pi}{8}+k\pi\left(k\in Z\right)\)

maxy=\(\sqrt{2}\Leftrightarrow sin\left(2x-\dfrac{\pi}{4}\right)=1\)\(\Leftrightarrow x=\dfrac{3\pi}{8}+k\pi\left(k\in Z\right)\)

c) \(y=3sinx+4cosx=5\left(\dfrac{3}{5}sinx+\dfrac{4}{5}cosx\right)\)

Đặt \(cosa=\dfrac{3}{5}\) và \(sina=\dfrac{4}{5}\)(vì cos²a+sin²a=1)

\(y=5\left(sinx.cosa+cosx.sina\right)\)\(=5sin\left(x+a\right)\)

\(\Rightarrow-5\le y\le5\)

miny=-5 <=> \(sin\left(x+a\right)=-1\)\(\Leftrightarrow x=-\dfrac{\pi}{2}-arc.sina+k2\pi\left(k\in Z\right)\)

maxy=5 <=> \(sin\left(x+a\right)=1\)\(\Leftrightarrow x=\dfrac{\pi}{2}-arc.sina+k2\pi\left(k\in Z\right)\)

(P/s1:cái x ở câu c ấy trông nó ngu ngu??
 P/s2:sau khi load lại câu hỏi ở 1 tab khác ,thấy 1 câu trả lời nhưng vẫn đăng vì cảm thấy bỏ đi hơi phí :?)

Áp dụng quy tắc sau: Nếu \(a\sin x+b\cos y=c\Leftrightarrow a^2+b^2\ge c^2\)

a/ \(3+1\ge y^2\Leftrightarrow4\ge y^2\Leftrightarrow-2\le y\le2\)

\(y_{max}=2\Leftrightarrow\sqrt{3}\sin x+\cos x=2\Leftrightarrow\dfrac{\sqrt{3}}{2}\sin x+\dfrac{1}{2}\cos x=1\Leftrightarrow\cos\dfrac{\pi}{6}.\sin x+\sin\dfrac{\pi}{6}.\cos x=1\)

\(\Rightarrow\sin\left(x+\dfrac{\pi}{6}\right)=1\Leftrightarrow x+\dfrac{\pi}{6}=\dfrac{\pi}{2}+k2\pi\Leftrightarrow x=\dfrac{\pi}{3}+k2\pi\)

\(y_{min}=-2\Leftrightarrow\sin\left(x+\dfrac{\pi}{6}\right)=-1\Leftrightarrow x+\dfrac{\pi}{6}=-\dfrac{\pi}{2}+k2\pi\Leftrightarrow x=-\dfrac{2}{3}\pi+k2\pi\)

\(y=sin2x+cos2x\)

Vì \(sinx,cosx\) xác định với mọi \(x\in R\) nên hàm số \(y=sin2x+cos2x\) xác định \(\forall x\in R\)

Vậy \(D=R\)

Đáp án: A

Ta có:

A = c o s 2 x + sin 2 x + sin 2 x 2 sin x + c o s x

Chọn A

y = cos6 x+ sin2xcos2x(sin2x + cos2x) + sin4x - sin2x

= cos6x + sin2x(1 - sin2x) + sin4x - sin2x = cos6x

Do đó : y' = -6cos5xsinx.

A =  cos 6 x  + 3 sin 2 x . cos 2 x  + 2 sin 4 α .  cos 2 x  +  sin 4 α

=  cos 6 x  + 3.(1 -  cos 2 x ) cos 4 x  + 2 sin 4 α . cos 2 x  +  sin 4 α

=  cos 6 x  + 3 cos 4 x  - 3 cos 6 x  + 2 sin 4 α  - 2 sin 6 x  +  sin 4 α

= -2.( cos 6 x  +  sin 6 x ) + 3  cos 4 x  + 3 sin 4 α

= -2.( cos 6 x  +  sin 6 x ) + 3.( cos 4 x  +  sin 4 α ) = 1

Vậy biểu thức A không phụ thuộc vào x.

Chọn A.

Từ giả  thiết suy ra:

A = (cos⁴x + cos²x sin²x) + sin²x = cos²x(sin²x + cos²x ) + sin²x

A = cos²x.1 + sin²x = 1

ta có : \(\dfrac{sinx-sin2x}{cosx+cos2x}=\dfrac{sinx-2sinxcosx}{cosx+2cos^2x-1}=\dfrac{-sinx\left(2cosx-1\right)}{cos^2x+cosx+cos^2x-1}\)

\(=\dfrac{-sinx\left(2cosx-1\right)}{cosx\left(cosx+1\right)+\left(cosx+1\right)\left(cosx-1\right)}=\dfrac{-sinx\left(2cosx-1\right)}{\left(2cosx-1\right)\left(cosx+1\right)}\)

\(=\dfrac{-sinx}{cosx+1}\)