a) \(4sinx-1=1\Leftrightarrow4sinx=2\Leftrightarrow sinx=\dfrac{2}{4}=\dfrac{1}{2}\)

\(\Leftrightarrow x=30^o\)

b) \(2\sqrt{3}-3tanx=\sqrt{3}\Leftrightarrow3tanx=2\sqrt{3}-\sqrt{3}=\sqrt{3}\Leftrightarrow tanx=\dfrac{\sqrt{3}}{3}\)

\(\Leftrightarrow x=30^o\)

c) \(7sinx-3cos\left(90^o-x\right)=2,5\Leftrightarrow7sinx-3sinx=2,5\Leftrightarrow4sinx=2,5\Leftrightarrow sinx=\dfrac{5}{8}\Leftrightarrow x=30^o41'\)

d)\(\left(2sin-\sqrt{2}\right)\left(4cos-5\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}2sin-\sqrt{2}=0\\4cos-5=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}2sin=\sqrt{2}\\4cos=5\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}sin=\dfrac{\sqrt{2}}{2}\\cos=\dfrac{5}{4}\left(loai\right)\end{matrix}\right.\)\(\Rightarrow x=45^o\)

Xin lỗi nãy đang làm thì bấm gửi, quên còn câu e, f nữa:"(

e) \(\dfrac{1}{cos^2x}-tanx=1\Leftrightarrow1+tan^2x-tanx-1=0\Leftrightarrow tan^2x-tanx=0\Leftrightarrow tanx\left(tanx-1\right)=0\Rightarrow tanx-1=0\Leftrightarrow tanx=1\Leftrightarrow x=45^o\)

f) \(cos^2x-3sin^2x=0,19\Leftrightarrow1-sin^2x-3sin^2x=0,19\Leftrightarrow1-4sin^2x=0,19\Leftrightarrow4sin^2x=0,81\Leftrightarrow sin^2x=\dfrac{81}{400}\Leftrightarrow sinx=\dfrac{9}{20}\Leftrightarrow x=26^o44'\)

1. 

ĐKXĐ: \(x\ne k\pi\)

\(\Leftrightarrow\left(2cos2x-1\right)\left(sinx-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cos2x=\dfrac{1}{2}\\sinx=3>1\left(ktm\right)\end{matrix}\right.\)

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

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

2. Bạn kiểm tra lại đề, pt này về cơ bản ko giải được.

3.

ĐKXĐ: \(x\ne\dfrac{k\pi}{2}\)

\(\dfrac{3\left(sinx+\dfrac{sinx}{cosx}\right)}{\dfrac{sinx}{cosx}-sinx}-2cosx=2\)

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

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

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

Vậy pt đã cho vô nghiệm

4.

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

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

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

\(\Leftrightarrow\sqrt{2}sin\left(x+\dfrac{\pi}{4}\right)\left(2cosx+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sin\left(x+\dfrac{\pi}{4}\right)=0\\cosx=-\dfrac{1}{2}\end{matrix}\right.\)

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

Bởi vì đơn giản cos luôn luôn nhỏ hơn 1

1 ) tanx+1/tanx =2 <=> tan^2x+1=tanx <=> (tanx-1)^2=0 <=> tanx=1 <=> x= pi/4+k.pi

có tanx.cotx=1 nên tanx=1/cotx. thay vào

Đề lỗi font. Bạn cần chỉnh sửa lại bằng công thức toán để được hỗ trợ tốt hơn.

Đề bị lỗi rồi bạn ơi

a) \(\left|sinx-cosx\right|+\left|sinx+cosx\right|=2\)

\(\Leftrightarrow\left(sinx-cosx\right)^2+2\left|sinx-cosx\right|\left|sinx+cosx\right|+\left(cosx+sinx\right)^2=4\)

\(\Leftrightarrow2\left(sin^2x+cos^2x\right)+2\left|\left(sinx-cosx\right)\left(sinx+cosx\right)\right|=4\)

\(\Leftrightarrow\left|sin^2x-cos^2x\right|=1\)

\(\Leftrightarrow\left[{}\begin{matrix}sin^2x-cos^2x=1\\sin^2x-cos^2x=-1\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}sin^2x-cos^2x=sin^2x+cos^2x\\sin^2x-cos^2x=-\left(sin^2x+cos^2x\right)\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}cos^2x=0\\sin^2x=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}cosx=0\\sinx=0\end{matrix}\right.\)\(\Rightarrow cosx.sinx=0\Rightarrow sin2x=0\)

\(\Rightarrow x=\dfrac{k\pi}{2},k\in Z\)

Vậy...

b) ĐK:\(x\ne\dfrac{k\pi}{2};k\in Z\)

Pt \(\Leftrightarrow\dfrac{sinx}{cosx}-\dfrac{3cosx}{sinx}=4\left(sinx+\sqrt{3}cosx\right)\)

\(\Leftrightarrow\dfrac{sin^2x-3cos^2x}{cosx.sinx}=4\left(sinx+\sqrt{3}cosx\right)\)

\(\Leftrightarrow\dfrac{\left(sinx-\sqrt{3}cosx\right)\left(sinx+\sqrt{3}cosx\right)}{sinx.cosx}=4\left(sinx+\sqrt{3}cosx\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx+\sqrt{3}cosx=0\left(1\right)\\\dfrac{sinx-\sqrt{3}cosx}{sinx.cosx}=4\left(2\right)\end{matrix}\right.\)

Từ \(\left(1\right)\Leftrightarrow tanx=-\sqrt{3}\Leftrightarrow x=-\dfrac{\pi}{3}+k\pi,k\in Z\)

Từ (2)\(\Leftrightarrow sinx-\sqrt{3}cosx=4sinx.cosx\)

\(\Leftrightarrow\dfrac{1}{2}sinx-\dfrac{\sqrt{3}}{2}cosx=2sinx.cosx\)

\(\Leftrightarrow sin\left(x-\dfrac{\pi}{3}\right)=sin2x\)\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{\pi}{3}+k2\pi\\x=\dfrac{4\pi}{9}+\dfrac{k2\pi}{3}\end{matrix}\right.\)\(\left(k\in Z\right)\)

Vậy \(\left[{}\begin{matrix}x=-\dfrac{\pi}{3}+k\pi\\x=\dfrac{4\pi}{9}+\dfrac{k2\pi}{3}\end{matrix}\right.\)\(\left(k\in Z\right)\)

c) ĐK: \(x\ne\dfrac{\pi}{4}+\dfrac{k\pi}{2}\left(k\in Z\right)\)

Pt \(\Leftrightarrow\left(\sqrt{2}sinx-1\right)^2+\left(\sqrt{3}tan2x-1\right)^2=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{2}sinx-1=0\\\sqrt{3}tan2x-1=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}sinx=\dfrac{1}{\sqrt{2}}\\tan2x=\dfrac{1}{\sqrt{3}}\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x=\dfrac{\pi}{4}+k2\pi\\x=\dfrac{3\pi}{4}+k2\pi\end{matrix}\right.\\x=\dfrac{\pi}{12}+k\pi\end{matrix}\right.\)\(\Rightarrow x\in\varnothing\)

Vậy pt vô nghiệm

Ta có sin²x + cos²x = 1 => sin²x = 1 - cos²x = 1 - (1/3)² = 8/9 => sinx = (2can2)/3

=> tanx = sinx/cosx = 2can2

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.

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

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

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

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

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

\(\Leftrightarrow cos2x=0\)

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