Với \(cosx=0\) ko phải nghiệm

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

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

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

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

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

câu này nhìn ngứa mắt quá làm kiểu gì giờ ??? 

c.

\(\Leftrightarrow cos\left(x+12^0\right)+cos\left(90^0-78^0+x\right)=1\)

\(\Leftrightarrow2cos\left(x+12^0\right)=1\)

\(\Leftrightarrow cos\left(x+12^0\right)=\dfrac{1}{2}\)

\(\Leftrightarrow\left[{}\begin{matrix}x+12^0=60^0+k360^0\\x+12^0=-60^0+k360^0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=48^0+k360^0\\x=-72^0+k360^0\end{matrix}\right.\)

2.

Do \(-1\le sin\left(3x-27^0\right)\le1\) nên pt có nghiệm khi:

\(\left\{{}\begin{matrix}2m^2+m\ge-1\\2m^2+m\le1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}2m^2+m+1\ge0\left(luôn-đúng\right)\\2m^2+m-1\le0\end{matrix}\right.\)

\(\Rightarrow-1\le m\le\dfrac{1}{2}\)

a.

\(\Rightarrow\left[{}\begin{matrix}x+15^0=arccos\left(\dfrac{2}{5}\right)+k360^0\\x+15^0=-arccos\left(\dfrac{2}{5}\right)+k360^0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-15^0+arccos\left(\dfrac{2}{5}\right)+k360^0\\x=-15^0-arccos\left(\dfrac{2}{5}\right)+k360^0\end{matrix}\right.\)

b.

\(2x-10^0=arccot\left(4\right)+k180^0\)

\(\Rightarrow x=5^0+\dfrac{1}{2}arccot\left(4\right)+k90^0\)

2.

Phương trình \(sin\left(3x-27^o\right)=2m^2+m\) có nghiệm khi:

\(2m^2+m\in\left[-1;1\right]\)

\(\Leftrightarrow\left\{{}\begin{matrix}2m^2+m\le1\\2m^2+m\ge-1\end{matrix}\right.\)

\(\Leftrightarrow\left(m+1\right)\left(2m-1\right)\le0\)

\(\Leftrightarrow-1\le m\le\dfrac{1}{2}\)

a) Dễ thấy cosx = 0 không thỏa mãn phương trình đã cho nên chiaw phương trình cho cos²x ta được phương trình tương đương 2tan²x + tanx - 3 = 0.

Đặt t = tanx thì phương trình này trở thành

2t² + t - 3 = 0 ⇔ t ∈ {1 ; }.

Vậy

b) Thay 2 = 2(sin²x + cos²x), phương trình đã cho trở thành

3sin²x - 4sinxcosx + 5cos²x = 2sin²x + 2cos²x

⇔ sin²x - 4sinxcosx + 3cos²x = 0

⇔ tan²x - 4tanx + 3 = 0

⇔

⇔ x = + kπ ; x = arctan3 + kπ, k ∈ Z.

c) Thay sin2x = 2sinxcosx ; = (sin²x + cos²x) vào phương trình đã cho và rút gọn ta được phương trình tương đương

sin²x + 2sinxcosx - cos²x = 0 ⇔ tan²x + 4tanx - 5 = 0 ⇔

⇔ x = + kπ ; x = arctan(-5) + kπ, k ∈ Z.

d) 2cos²x - 3√3sin2x - 4sin²x = -4

⇔ 2cos²x - 3√3sin2x + 4 - 4sin²x = 0

⇔ 6cos²x - 6√3sinxcosx = 0 ⇔ cosx(cosx - √3sinx) = 0

⇔

a.

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

\(\Leftrightarrow2-\left(2sin\dfrac{x}{2}cos\dfrac{x}{2}\right)^2=1\)

\(\Leftrightarrow1-sin^2x=0\)

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

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

b.

\(\left(sin^2x+cos^2x\right)^3-3sin^2x.cos^2x\left(sin^2x+cos^2x\right)=\dfrac{7}{16}\)

\(\Leftrightarrow1-\dfrac{3}{4}\left(2sinx.cosx\right)^2=\dfrac{7}{16}\)

\(\Leftrightarrow16-12.sin^22x=7\)

\(\Leftrightarrow3-4sin^22x=0\)

\(\Leftrightarrow3-2\left(1-cos4x\right)=0\)

\(\Leftrightarrow cos4x=-\dfrac{1}{2}\)

\(\Leftrightarrow4x=\pm\dfrac{2\pi}{3}+k2\pi\)

\(\Leftrightarrow x=\pm\dfrac{\pi}{6}+\dfrac{k\pi}{2}\)

c.

\(\left(sin^2x+cos^2x\right)^3-3sin^2x.cos^2x\left(sin^2x+cos^2x\right)=cos^22x+\dfrac{1}{4}\)

\(\Leftrightarrow1-\dfrac{3}{4}\left(2sinx.cosx\right)^2=cos^22x+\dfrac{1}{4}\)

\(\Leftrightarrow3-3sin^22x=4cos^22x\)

\(\Leftrightarrow3=3\left(sin^22x+cos^22x\right)+cos^22x\)

\(\Leftrightarrow3=3+cos^22x\)

\(\Leftrightarrow cos2x=0\)

\(\Leftrightarrow x=\dfrac{\pi}{4}+\dfrac{k\pi}{2}\)

a)Đặt \(t=sinx+cosx\);\(t\in\left[-\sqrt{2};\sqrt{2}\right]\)

\(\Leftrightarrow t^2=sin^2+2sinx.cosx+cos^2x\)

\(\Leftrightarrow t^2=1+2sinx.cosx\)

\(\Leftrightarrow\dfrac{t^2-1}{2}=sinx.cosx\)

Pttt: \(3t-4.\dfrac{t^2-1}{2}=0\) \(\Leftrightarrow-2t^2+3t+2=0\)

\(\Leftrightarrow\left[{}\begin{matrix}t=2\left(ktm\right)\\t=-\dfrac{1}{2}\left(tm\right)\end{matrix}\right.\)

\(\Rightarrow sinx.cosx=-\dfrac{3}{8}\) \(\Leftrightarrow2sinx.cosx=-\dfrac{3}{4}\)\(\Leftrightarrow sin2x=-\dfrac{3}{4}\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}.arc.sin\left(-\dfrac{3}{4}\right)+k\pi\\x=\dfrac{\pi}{2}-\dfrac{1}{2}.arc.sin\left(-\dfrac{3}{4}\right)+k\pi\end{matrix}\right.\), \(k\in Z\)

Vậy...

b)Pt 

Đặt \(t=sinx-cosx;t\in\left[-\sqrt{2};\sqrt{2}\right]\)

\(\Leftrightarrow t^2-1=-2sinx.cosx\)

Pttt:\(12t+t^2-1=2\)

\(\Leftrightarrow\left[{}\begin{matrix}t=-6+\sqrt{39}\left(tm\right)\\t=-6-\sqrt{39}\left(ktm\right)\end{matrix}\right.\)

\(\Rightarrow cosx+sinx=-6+\sqrt{39}\)

\(\Leftrightarrow\sqrt{2}.cos\left(x-\dfrac{\pi}{4}\right)=-6+\sqrt{39}\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{4}+arc.cos\left(\dfrac{-6+\sqrt{39}}{\sqrt{2}}\right)+k2\pi\\x=\dfrac{\pi}{4}-arc.cos\left(\dfrac{-6+\sqrt{39}}{2}\right)+k2\pi\end{matrix}\right.\)\(,k\in Z\)

Vậy...(Nghiệm xấu)