1+tan^2 x = 1/cos^2 x

=> 1+ t^2 = 1/cos^2 x

=> 3 + 3t^2 = 3/cos^2 x

PT TRỞ THÀNH :

3 + 3t^2 - 2t + 1 = 0

<=> 3t^2 - 2t + 4 = 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.

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

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

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

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

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

\(\Leftrightarrow cos2x=0\)

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

1.

\(0< x< \dfrac{\pi}{2}\Rightarrow cosx>0\)

\(\Rightarrow cosx=\sqrt{1-sin^2x}=\dfrac{\sqrt{5}}{3}\)

\(tanx=\dfrac{sinx}{cosx}=\dfrac{2}{\sqrt{5}}\)

\(sin\left(x+\dfrac{\pi}{4}\right)=\dfrac{\sqrt{2}}{2}\left(sinx+cosx\right)=\dfrac{\sqrt{10}+2\sqrt{2}}{6}\)

2.

Đề bài thiếu, cos?x

Và x thuộc khoảng nào?

3.

\(x\in\left(0;\dfrac{\pi}{2}\right)\Rightarrow sinx;cosx>0\)

\(\dfrac{1}{cos^2x}=1+tan^2x=5\Rightarrow cos^2x=\dfrac{1}{5}\Rightarrow cosx=\dfrac{\sqrt{5}}{5}\)

\(sinx=cosx.tanx=\dfrac{2\sqrt{5}}{5}\)

4.

\(A=\left(2cos^2x-1\right)-2cos^2x+sinx+1=sinx\)

\(B=\dfrac{cos3x+cosx+cos2x}{cos2x}=\dfrac{2cos2x.cosx+cos2x}{cos2x}=\dfrac{cos2x\left(2cosx+1\right)}{cos2x}=2cosx+1\)

Bạn tham khảo nha. Chúc bạn học tốt

Pt vô nghiệm

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

ĐK: \(x\ne-\dfrac{\pi}{4}+k\pi\)

\(\dfrac{tanx}{1-tan^2x}=\dfrac{1}{2}cot\left(x+\dfrac{\pi}{4}\right)\)

\(\Leftrightarrow\dfrac{2tanx}{1-tan^2x}=tan\left(\dfrac{\pi}{4}-x\right)\)

\(\Leftrightarrow tan2x=tan\left(\dfrac{\pi}{4}-x\right)\)

\(\Leftrightarrow2x=\dfrac{\pi}{4}-x+k\pi\)

\(\Leftrightarrow x=\dfrac{\pi}{12}+\dfrac{k\pi}{3}\)

Điều kiện:

⇔ tan x.(1 - tanx) + tanx + 1 = 1 – tan x.

⇔ tan x - tan2x + 2.tan x = 0

⇔ tan2x - 3tanx = 0

⇔ tanx(tanx - 3) = 0

Vậy phương trình đã cho có tập nghiệm là:

{arctan 3+kπ; k ∈ Z }

a) cos3x =  \(cos\left(\pi-x-\dfrac{\pi}{3}\right)\)

<=> cos3x = \(cos\left(\dfrac{2\pi}{3}-x\right)\)

<=> 3x = \(\dfrac{2\pi}{3}-x\) hoặc 3x = \(\dfrac{-2\pi}{3}+x\)

<=> 4x = \(\dfrac{2\pi}{3}+k2\pi\) hoặc 2x = \(\dfrac{-2\pi}{3}+k2\pi\)

<=> x = \(\dfrac{\pi}{6}+\dfrac{k\pi}{2}\) hoặc x = \(\dfrac{-\pi}{3}+k\pi\)

<=> x = \(\left\{\dfrac{\pi}{6}+\dfrac{k\pi}{2};\dfrac{-\pi}{3}+k\pi;k\in Z\right\}\)

b ) Điều kiện sinx\(\ne0;cosx\ne0\)

<=> sin2x\(\ne0\) <=> x \(\ne\dfrac{k\pi}{2}\);k\(\in Z\)

tanx + cotx =0

<=> tan²x + tanx =0

<=> tanx(tanx+1)=0

<=> tanx=0 hoặc tanx = -1

<=> x=\(k\pi\) (loại) hoặc x = \(\dfrac{-\pi}{4}+k\pi\)

Vậy x = \(\dfrac{-\pi}{4}+k\pi;k\in Z\)