- Ta có :

+) Xét (I) ta có:

+) Do đó, (I) đúng và (II) sai.

Chọn B.

- Ta có :

+) Xét (I) ta có:

+) Do đó, (I) đúng và (II) sai.

Chọn B. 

b:

ĐKXĐ: \(\left\{{}\begin{matrix}cosx< >0\\sinx< >0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x< >\dfrac{\Omega}{2}+k\Omega\\x\ne k\Omega\end{matrix}\right.\)

=>\(x\ne\dfrac{\Omega}{2}+\dfrac{k\Omega}{2}\)

 \(\dfrac{1}{cosx}+\dfrac{\sqrt{3}}{sinx}=2\cdot sin\left(x+\dfrac{\Omega}{3}\right)\)

=>\(\dfrac{sinx+\sqrt{3}\cdot cosx}{cosx\cdot sinx}=2\cdot sin\left(x+\dfrac{\Omega}{3}\right)\)

=>\(\dfrac{sinx+\sqrt{3}\cdot cosx}{cosx\cdot sinx}=2\cdot\left[sinx\cdot\cos\dfrac{\Omega}{3}+sin\left(\dfrac{\Omega}{3}\right)\cdot cosx\right]\)

=>\(\dfrac{sinx+\sqrt{3}\cdot cosx}{cosx\cdot sinx}=2\cdot\left(\dfrac{1}{2}\cdot sinx+\dfrac{\sqrt{3}}{2}\cdot cosx\right)\)

=>\(\left(sinx+\sqrt{3}\cdot cosx\right)\left(\dfrac{1}{cosx\cdot sinx}-1\right)=0\)

=>\(2\cdot\left(sinx\cdot\dfrac{1}{2}+\dfrac{\sqrt{3}}{2}\cdot cosx\right)\cdot\left(\dfrac{2}{2\cdot sinx\cdot cosx}-1\right)=0\)

=>\(2\cdot sin\left(x+\dfrac{\Omega}{3}\right)\cdot\left(\dfrac{2}{sin2x}-1\right)=0\)

=>\(\left[{}\begin{matrix}sin\left(x+\dfrac{\Omega}{3}\right)=0\\\dfrac{2}{sin2x}-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{\Omega}{3}=k\Omega\\sin2x=2\left(loại\right)\end{matrix}\right.\)

=>\(x=-\dfrac{\Omega}{3}+k\Omega\)

?

V?

\(\left(\sin x+\cos x\right)^2=\dfrac{9}{16}\)

nên \(1+2\cdot\sin x\cdot\cos x=\dfrac{9}{16}\)

\(\Leftrightarrow\sin x\cdot\cos x=-\dfrac{7}{32}\)

\(\left|\sin x-\cos x\right|=\sqrt{\left(\sin x+\cos x\right)^2-4\cdot\sin x\cdot\cos x}\)

\(=\sqrt{\dfrac{9}{16}-4\cdot\dfrac{-7}{32}}=\dfrac{\sqrt{23}}{4}\)

nên \(\left[{}\begin{matrix}\sin x-\cos x=\dfrac{\sqrt[4]{23}}{2}\\\sin x-\cos x=-\dfrac{\sqrt[4]{23}}{2}\end{matrix}\right.\)

Chọn đáp án B.

Đáp án: B.

Hướng dẫn: Đặt u = (x + 1), v' = sinx.

Đáp án: B.

Hướng dẫn: Đặt u = (x + 1), v' = sinx.

1.

\(\Leftrightarrow2sin\frac{x}{2}cos\frac{x}{2}+\sqrt{3}cos\frac{x}{2}=0\)

\(\Leftrightarrow cos\frac{x}{2}\left(2sin\frac{x}{2}+\sqrt{3}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cos\frac{x}{2}=0\\sin\frac{x}{2}=-\frac{\sqrt{3}}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\frac{x}{2}=\frac{\pi}{2}+k\pi\\\frac{x}{2}=-\frac{\pi}{3}+k2\pi\\\frac{x}{2}=\frac{4\pi}{3}+k2\pi\end{matrix}\right.\) \(\Leftrightarrow...\)

2.

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

\(\Leftrightarrow cosx=cos\left(x-\frac{\pi}{3}\right)\)

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

\(\Rightarrow x=\frac{\pi}{6}+k\pi\)

3.

\(\Leftrightarrow\frac{1}{2}sinx-\frac{\sqrt{3}}{2}cosx=0\)

\(\Leftrightarrow sin\left(x-\frac{\pi}{3}\right)=0\)

\(\Leftrightarrow x-\frac{\pi}{3}=k\pi\)

\(\Leftrightarrow...\)

4.

\(1+\frac{1}{2}sin6x=sin^2x+cos^2x+2sinx.cosx\)

\(\Leftrightarrow\frac{1}{2}sin6x=sin2x\)

\(\Leftrightarrow sin6x-2sin2x=0\)

\(\Leftrightarrow3sin2x-4sin^32x-2sin2x=0\)

\(\Leftrightarrow sin2x-4sin^32x=0\)

\(\Leftrightarrow sin2x\left(1-4sin^22x\right)=0\)

\(\Leftrightarrow sin2x\left(2cos2x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sin2x=0\\cos2x=\frac{1}{2}\end{matrix}\right.\) \(\Leftrightarrow...\)