\(\sqrt{3}cosx+2sin^2\left(\dfrac{x}{2}-\pi\right)=1\) 

\(\Leftrightarrow\sqrt{3}cosx+2sin^2\dfrac{x}{2}=1\)

\(\Leftrightarrow\sqrt{3}cosx-cosx=0\Leftrightarrow cosx=0\Leftrightarrow x=\dfrac{\pi}{2}+k\pi\) ( k thuộc Z )

Vậy ... 

22.

Nhận thấy \(cosx=0\) không phải nghiệm, chia 2 vế cho \(cos^2x\)

\(3tan^2x+2tanx-1=0\)

\(\Leftrightarrow\left[{}\begin{matrix}tanx=-1\\tanx=\dfrac{1}{3}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{\pi}{4}+k\pi\\x=arctan\left(\dfrac{1}{3}\right)+k\pi\end{matrix}\right.\)

Nghiệm dương nhỏ nhất của pt là: \(x=arctan\left(\dfrac{1}{3}\right)\)

22. PT đã cho tương đương

3 - 4cos²x + 2 sinxcosx = 0

⇔ 3 - 2 - 2cos2x + sin2x = 0

⇔ 1 - 2cos2x + sin2x = 0

⇔ 1 + sin2x = 2cos2x

⇔ sin\(\dfrac{\pi}{2}\) + sin2x = 2cos2x

⇔ \(2sin\left(\dfrac{\pi}{4}+x\right).cos\left(\dfrac{\pi}{4}-x\right)\) = 2cos2x

Do \(\left(\dfrac{\pi}{4}-x\right)+\left(\dfrac{\pi}{4}+x\right)=\dfrac{\pi}{2}\) 

⇒ \(sin\left(\dfrac{\pi}{4}+x\right)=cos\left(\dfrac{\pi}{4}-x\right)\)

Vậy sin²\(\left(x+\dfrac{\pi}{4}\right)\) = cos2x

Cái này là hiển nhiên ????

Pt \(\Leftrightarrow sin^2x+2.sinx.cosx-2cos^2x=\dfrac{1}{2}\left(sin^2x+cos^2x\right)\)

\(\Leftrightarrow sin^2x.\dfrac{1}{2}+2.sinx.cosx-\dfrac{5}{2}cos^2x=0\)

\(\Leftrightarrow\left(sinx-cosx\right)\left(sinx+5cosx\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx=cosx\\sinx=-5cosx\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}tanx=1\\tanx=-5\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{4}+k\pi\\x=arc.tan\left(-5\right)+k\pi\end{matrix}\right.\)(\(k\in Z\))

Vậy...

ĐKXĐ: \(\left\{{}\begin{matrix}x\ne-\dfrac{\pi}{2}+k2\pi\\x\ne\dfrac{\pi}{6}+\dfrac{k2\pi}{3}\\\end{matrix}\right.\)

\(\dfrac{cosx-2sinx.cosx}{2cos^2x-1-sinx}=\sqrt{3}\)

\(\Leftrightarrow\dfrac{cosx-sin2x}{cos2x-sinx}=\sqrt{3}\)

\(\Rightarrow cosx-sin2x=\sqrt{3}cos2x-\sqrt{3}sinx\)

\(\Leftrightarrow cosx+\sqrt{3}sinx=\sqrt{3}cos2x+sin2x\)

\(\Leftrightarrow\dfrac{1}{2}cosx+\dfrac{\sqrt{3}}{2}sinx=\dfrac{\sqrt{3}}{2}cos2x+\dfrac{1}{2}sin2x\)

\(\Leftrightarrow cos\left(x-\dfrac{\pi}{3}\right)=cos\left(2x-\dfrac{\pi}{6}\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-\dfrac{\pi}{6}=x-\dfrac{\pi}{3}+k2\pi\\2x-\dfrac{\pi}{6}=\dfrac{\pi}{3}-x+k2\pi\end{matrix}\right.\)

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

Vậy \(x=-\dfrac{\pi}{6}+k2\pi\)

a.

ĐKXĐ: \(x\ne\dfrac{\pi}{2}+k\pi\)

Chia 2 vế cho cosx:

\(tanx+1=\dfrac{1}{cos^2x}\)

\(\Rightarrow tanx+1=1+tan^2x\)

\(\Rightarrow\left[{}\begin{matrix}tanx=0\\tanx=1\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=k\pi\\x=\dfrac{\pi}{4}+k\pi\end{matrix}\right.\)

c.

\(\Leftrightarrow2sin2x+2sin^2x=1\)

\(\Leftrightarrow2sin2x=1-2sin^2x\)

\(\Leftrightarrow2sin2x=cos2x\)

\(\Rightarrow tan2x=\dfrac{1}{2}\)

\(\Rightarrow2x=arctan\left(\dfrac{1}{2}\right)+k\pi\)

\(\Rightarrow x=\dfrac{1}{2}arctan\left(\dfrac{1}{2}\right)+\dfrac{k\pi}{2}\)

b.

\(\Leftrightarrow4sin2x+3sin\left(\dfrac{\pi}{2}-2x\right)=5\)

\(\Leftrightarrow4sin2x+3cos2x=5\)

\(\Leftrightarrow\dfrac{4}{5}sin2x+\dfrac{3}{5}cos2x=1\)

Đặt \(\dfrac{4}{5}=cosa\) với \(a\in\left(0;\dfrac{\pi}{2}\right)\Rightarrow\dfrac{3}{5}=sina\)

\(\Rightarrow sin2x.cosa+cos2x.sina=1\)

\(\Rightarrow sin\left(2x+a\right)=1\)

\(\Rightarrow2x+a=\dfrac{\pi}{2}+k2\pi\)

\(\Rightarrow x=-\dfrac{a}{2}+\dfrac{\pi}{4}+k\pi\)

\(\left(2x+1\right)\left(2x+3\right)\left(x+1\right)^2=18\)

\(\Leftrightarrow\left(2x+2-1\right)\left(2x+2+1\right)\left(x+1\right)^2=18\)

\(\Leftrightarrow\left(\left(2x+2\right)^2-1\right)\left(x+1\right)^2=18\)

\(\Leftrightarrow4\left(x+1\right)^4-\left(x+1\right)^2-18=0\)

Đặt t = \(\left(x+1\right)^2\) \(\left(t\ge0\right)\)

pt \(\Leftrightarrow4t^2-t-18=0\)

\(\Leftrightarrow\left[{}\begin{matrix}t=\dfrac{9}{4}\left(nh\right)\\t=-2\left(l\right)\end{matrix}\right.\)

\(\Leftrightarrow\left(x+1\right)^2-\dfrac{9}{4}=0\)

\(\Leftrightarrow\left(x+1-\dfrac{3}{2}\right)\left(x+1+\dfrac{3}{2}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=\dfrac{5}{2}\end{matrix}\right.\)

ĐKXĐ: \(\left\{{}\begin{matrix}x\ge1\\-4+\sqrt{7}\le x\le-1\end{matrix}\right.\)

Khi x thỏa ĐKXĐ, vế phải luôn dương, bình phương 2 vế ta được:

\(\Leftrightarrow3x^2+16x+17+2\sqrt{\left(x^2-1\right)\left(2x^2+16x+18\right)}=4x^2+16x+16\)

\(\Leftrightarrow2\sqrt{\left(x^2-1\right)\left(2x^2+16x+18\right)}=x^2-1\)

\(\Leftrightarrow4\left(x^2-1\right)\left(2x^2+16x+18\right)=\left(x^2-1\right)^2\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2-1=0\\4\left(2x^2+16x+18\right)=x^2-1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\pm1\\7x^2+64x+73=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\pm1\\x=\dfrac{-32+3\sqrt{57}}{7}\\x=\dfrac{-32-3\sqrt{57}}{7}\left(loại\right)\end{matrix}\right.\)

(2x+1)(x+1)²(2x+3)=18 

<=> (2x+2-1)(x+1)²(2x+2+1)=18

Đặt y=x+1, ta có: 

(2y-1)y²(2y+1)=18

Ta có 

(2x+1)(x+1)²(2x+3)=18

=> (x+1)²(4x²+8x+3)-18=0

=> (x²+2x+1)(4x²+8x+3)-18=0

Đặt x²+2x+1=a ta có 

a.(4a-1)-18=0

=> 4a²-a-18=0

=> 4a² +8a-9a-18=0

=> 4a(a+2)-9(a+2)=0

=> (a+2)(4a-9)=0

Với a=x²+2x+1biểu thức trên trở thành

(x²+2x+3)(4x²+8x-5)=0

=> x²+2x+3=0 hoặc 4x²+8x-5=0

• x²+2x+3=0 => phương trình vô nghiệm

• 4x²+8x-5=0 => x=1/2 hoặc x=-5/2

Vậy x=1/2 và x=-5/2 là nghiệm của phương trình

a, Ta có : \(\sin\left(3x+60\right)=\dfrac{1}{2}\)

\(\Rightarrow3x+60=30+2k180\)

\(\Rightarrow3x=2k180-30\)

\(\Leftrightarrow x=120k-10\)

Vậy ...

b, Ta có : \(\cos\left(2x-\dfrac{\pi}{3}\right)=-\dfrac{\sqrt{2}}{2}\)

\(\Rightarrow2x-\dfrac{\pi}{3}=\dfrac{3}{4}\pi+k2\pi\)

\(\Leftrightarrow x=\dfrac{13}{24}\pi+k\pi\)

Vậy ...

c, Ta có : \(tan\left(x+\dfrac{\pi}{6}\right)=\sqrt{3}\)

\(\Rightarrow x+\dfrac{\pi}{6}=\dfrac{\pi}{3}+k\pi\)

\(\Leftrightarrow x=\dfrac{\pi}{6}+k\pi\)

Vậy ...

d, Ta có : \(\cot\left(2x+\pi\right)=-1\)

\(\Rightarrow2x+\pi=\dfrac{3}{4}\pi+k\pi\)

\(\Leftrightarrow x=-\dfrac{1}{8}\pi+\dfrac{k}{2}\pi\)

Vậy ...

a) \(sin\left(3x+60^0\right)=\dfrac{1}{2}\)

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

\(\Leftrightarrow\left[{}\begin{matrix}3x+\dfrac{\pi}{3}=\dfrac{\pi}{6}+k2\pi\\3x+\dfrac{\pi}{3}=\dfrac{5\pi}{6}+k2\pi\end{matrix}\right.\)(\(k\in Z\))\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-\pi}{18}+\dfrac{k2\pi}{3}\\x=\dfrac{\pi}{6}+\dfrac{k2\pi}{3}\end{matrix}\right.\)(\(k\in Z\))

Vậy...

b) Pt\(\Leftrightarrow cos\left(2x-\dfrac{\pi}{3}\right)=cos\dfrac{3\pi}{4}\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-\dfrac{\pi}{3}=\dfrac{3\pi}{4}+k2\pi\\2x-\dfrac{\pi}{3}=-\dfrac{3\pi}{4}+k2\pi\end{matrix}\right.\)(\(k\in Z\))\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{13\pi}{24}+k\pi\\x=-\dfrac{5\pi}{24}+k\pi\end{matrix}\right.\)(\(k\in Z\))

Vậy...

c) Pt \(\Leftrightarrow tan\left(x+\dfrac{\pi}{6}\right)=tan\dfrac{\pi}{3}\)

\(\Leftrightarrow x+\dfrac{\pi}{6}=\dfrac{\pi}{3}+k\pi,k\in Z\)\(\Leftrightarrow x=\dfrac{\pi}{6}+k\pi,k\in Z\)

Vậy...

d) Pt \(\Leftrightarrow tan\left(2x+\pi\right)=-1\)

\(\Leftrightarrow2x+\pi=-\dfrac{\pi}{4}+k\pi,k\in Z\)

\(\Leftrightarrow x=-\dfrac{5\pi}{8}+\dfrac{k\pi}{2},k\in Z\)

Vậy...