Chọn C

Ta có: nên (1) và (2) có nghiệm.

Cách 1:

Xét: nên (3) vô nghiệm.

Cách 2:

Điều kiện có nghiệm của phương trình: sin x + cos x = 2 là:

(vô lý) nên (3) vô nghiệm.

Cách 3:

Vì 

nên (3) vô nghiệm.

1: \(\sin\left(x+\frac{\pi}{4}\right)=\frac23\)

=>\(\left[\begin{array}{l}x+\frac{\pi}{4}=\arcsin\left(\frac23\right)+k2\pi\\ x+\frac{\pi}{4}=\pi-\arcsin\left(\frac23\right)+k2\pi\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\arcsin\left(\frac23\right)-\frac{\pi}{4}+k2\pi\\ x=\frac34\pi-\arcsin\left(\frac23\right)+k2\pi\end{array}\right.\)

2: \(cos2x-5\cdot\sin x-3=0\)

=>\(1-2\cdot\sin^2x-5\cdot\sin x-3=0\)

=>\(-2\cdot\sin^2x-5\cdot\sin x-2=0\)

=>\(2\cdot\sin^2x+5\cdot\sin x+2=0\)

=>(sin x+2)(2 sin x+1)=0

=>2 sin x+1=0

=>sin x=-1/2

=>\(\left[\begin{array}{l}x=-\frac{\pi}{6}+k2\pi\\ x=\pi+\frac{\pi}{6}+k2\pi=\frac76\pi+k2\pi\end{array}\right.\)

3: \(cos3x=\sin2x\)

=>\(cos3x=cos\left(\frac{\pi}{2}-2x\right)\)

=>\(\left[\begin{array}{l}3x=\frac{\pi}{2}-2x+k2\pi\\ 3x=2x-\frac{\pi}{2}+k2\pi\end{array}\right.\Rightarrow\left[\begin{array}{l}5x=\frac{\pi}{2}+k2\pi\\ x=-\frac{\pi}{2}+k2\pi\end{array}\right.\)

=>\(\left[\begin{array}{l}x=\frac{\pi}{10}+\frac{k2\pi}{5}\\ x=-\frac{\pi}{2}+k2\pi\end{array}\right.\)

đáp án không giống lắm 

Dạ em cảm ơn ạ

a)pt\(\Leftrightarrow cosx\left(cosx+1\right)+sinx.sin^2x=0\)

\(\Leftrightarrow cosx\left(cosx+1\right)+sinx\left(1-cos^2x\right)=0\)

\(\Leftrightarrow\left(cosx+1\right)\left(cosx+sinx-sinx.cosx\right)=0\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}cosx=1\Leftrightarrow x=\pi+k2\pi\\cosx+sinx-sinx.cosx=0\left(\cdot\right)\end{array}\right.\)

Xét pt(*):

Đặt \(t=cosx+sinx,t\in\left[-\sqrt{2};\sqrt{2}\right]\Rightarrow sinx.cosx=\frac{t^2-1}{2}\)

(*) trở thành:\(t^2-2t-1=0\Leftrightarrow\left[\begin{array}{nghiempt}t=1-\sqrt{2}\\t=1+\sqrt{2}\left(L\right)\end{array}\right.\)

+)\(t=1-\sqrt{2}\Rightarrow\sqrt{2}sin\left(x+\frac{\pi}{4}\right)=1-\sqrt{2}\\ \Leftrightarrow\left[\begin{array}{nghiempt}x=-\frac{\pi}{4}+arcsin\left(\frac{-2+\sqrt{2}}{2}\right)+k2\pi\\x=-\frac{5\pi}{4}-arcsin\left(\frac{-2+\sqrt{2}}{2}\right)+k2\pi\end{cases}\left(k\in Z\right)}\)

a: tan x(cot^2x-1)

\(=\dfrac{1}{cotx}\left(cot^2x-cotx\cdot tanx\right)\)

=cotx-tanx/cotx=cotx(1-tan^2x)

b: \(tan^2x-sin^2x=\dfrac{sin^2x}{cos^2x}-sin^2x\)

\(=sin^2x\left(\dfrac{1}{cos^2x}-1\right)=sin^2x\cdot\dfrac{sin^2x}{cos^2x}=sin^2x\cdot tan^2x\)

c: \(\dfrac{cos^2x-sin^2x}{cot^2x-tan^2x}=\dfrac{cos^2x-sin^2x}{\dfrac{cos^2x}{sin^2x}-\dfrac{sin^2x}{cos^2x}}\)

\(=\left(cos^2x-sin^2x\right):\dfrac{cos^4x-sin^4x}{sin^2x\cdot cos^2x}\)

\(=\dfrac{sin^2x\cdot cos^2x}{1}=sin^2x\cdot cos^2x\)

=>sin^2x*cos^2x-cos^2x=cos^2x(sin^2x-1)

=-cos^2x*cos^2x=-cos^4x

=>ĐPCM

Với \(cosx=0\) ko phải nghiệm

Với \(cosx\ne0\) chia 2 vế cho \(cos^2x\)

\(\Rightarrow tan^2x-4\sqrt{3}tanx+1=-2\left(1+tan^2x\right)\)

\(\Leftrightarrow3tan^2x-4\sqrt{3}tanx+3=0\)

\(\Rightarrow\left[{}\begin{matrix}tanx=\sqrt{3}\\tanx=\dfrac{\sqrt{3}}{3}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{3}+k\pi\\x=\dfrac{\pi}{6}+k\pi\end{matrix}\right.\)