a. ĐKXĐ: ...

\(\frac{sinx}{cosx}+\frac{sin2x}{cos2x}+\frac{sin3x}{cos3x}=0\)

\(\Leftrightarrow\frac{sin2x.cosx+cos2x.sinx}{cosx.cos2x}+\frac{sin3x}{cos3x}=0\)

\(\Leftrightarrow\frac{sin3x}{cosx.cos2x}+\frac{sin3x}{cos3x}=0\)

\(\Leftrightarrow sin3x\left(\frac{cosx.cos2x+cos3x}{cosx.cos2x.cos3x}\right)=0\)

\(\Leftrightarrow sin3x\left(\frac{cosx\left(2cos^2x-1\right)+4cos^3x-3cosx}{cosx.cos2x.cos3x}\right)=0\)

\(\Leftrightarrow sin3x\left(\frac{6cos^2x-4}{cos2x.cos3x}\right)=0\)

\(\Leftrightarrow sin3x\left(\frac{3cos2x-1}{cos2x.cos3x}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sin3x=0\\cos2x=\frac{1}{3}\end{matrix}\right.\)

b.

\(cos2x\left(2cos^22x-1\right)=\frac{1}{2}\)

\(\Leftrightarrow4cos^32x-2cos2x-1=0\)

Pt bậc 3 này ko giải được, chắc bạn ghi nhầm đề

c. ĐKXĐ: ...

\(\frac{cosx}{sinx}-\frac{sinx}{cosx}=cosx-sinx\)

\(\Leftrightarrow\frac{\left(cosx-sinx\right)\left(cosx+sinx\right)}{sinx.cosx}=cosx-sinx\)

\(\Leftrightarrow\left[{}\begin{matrix}cosx-sinx=0\Rightarrow x=...\\\frac{cosx+sinx}{sinx.cosx}=1\left(1\right)\end{matrix}\right.\)

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

Đặt \(sinx+cosx=t\Rightarrow\left\{{}\begin{matrix}\left|t\right|\le\sqrt{2}\\sinx.cosx=\frac{t^2-1}{2}\end{matrix}\right.\)

\(\Rightarrow t=\frac{t^2-1}{2}\Rightarrow t^2-2t-1=0\Rightarrow\left[{}\begin{matrix}t=1+\sqrt{2}\left(l\right)\\t=1-\sqrt{2}\end{matrix}\right.\)

\(\Rightarrow\sqrt{2}sin\left(x+\frac{\pi}{4}\right)=1-\sqrt{2}\Rightarrow sin\left(x+\frac{\pi}{4}\right)=\frac{1-\sqrt{2}}{\sqrt{2}}\Rightarrow x=...\)

d.

\(\Leftrightarrow2\left(cosx-cos3x\right)sin4x=sin3x\)

\(\Leftrightarrow2sin4x.cosx-2sin4x.cos3x=sin3x\)

\(\Leftrightarrow sin5x+sin3x-sin7x-sinx=sin3x\)

\(\Leftrightarrow sin5x-sin7x-sinx=0\)

\(\Leftrightarrow-2cos6x.sinx-sinx=0\)

\(\Leftrightarrow sinx\left(2cos6x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx=0\\cos6x=-\frac{1}{2}\end{matrix}\right.\)

\(\Leftrightarrow...\)

tham khảogiúp mình nhé: (tanx + cotx)^2 - (tanx + cotx) = 2? | Yahoo Hỏi & Đáp

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:

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

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

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

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

ĐKXĐ: \(x\ne\dfrac{k\pi}{2}\)

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

\(\Leftrightarrow\dfrac{cosx-sinx}{sinx}=cosx\left(cosx-sinx\right)-sinx\left(cosx-sinx\right)\)

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

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

\(\Leftrightarrow\left(cosx-sinx\right)\left(3-sin2x-cos2x\right)=0\)

\(\Leftrightarrow\left(cosx-sinx\right)\left(3-\sqrt{2}sin\left(2x+\dfrac{\pi}{4}\right)\right)=0\)

mai đăng lại bài này nhé t làm cho h đi ngủ

a) \(\sin x = \frac{{\sqrt 3 }}{2}\;\; \Leftrightarrow \sin x = \sin \frac{\pi }{3}\;\;\; \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{x = \frac{\pi }{3} + k2\pi }\\{x = \pi  - \frac{\pi }{3} + k2\pi }\end{array}} \right.\;\;\; \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{x = \frac{\pi }{3} + k2\pi }\\{x = \frac{{2\pi }}{3} + k2\pi \;}\end{array}\;} \right.\left( {k \in \mathbb{Z}} \right)\)

b) \(2\cos x =  - \sqrt 2 \;\; \Leftrightarrow \cos x =  - \frac{{\sqrt 2 }}{2}\;\;\; \Leftrightarrow \cos x = \cos \frac{{3\pi }}{4}\;\;\; \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{x = \frac{{3\pi }}{4} + k2\pi }\\{x =  - \frac{{3\pi }}{4} + k2\pi }\end{array}\;\;\left( {k \in \mathbb{Z}} \right)} \right.\)

c) \(\sqrt 3 \;\left( {\tan \frac{x}{2} + {{15}^0}} \right) = 1\;\;\; \Leftrightarrow \tan \left( {\frac{x}{2} + \frac{\pi }{{12}}} \right) = \frac{1}{{\sqrt 3 }}\;\; \Leftrightarrow \tan \left( {\frac{x}{2} + \frac{\pi }{{12}}} \right) = \tan \frac{\pi }{6}\)

\( \Leftrightarrow \frac{x}{2} + \frac{\pi }{{12}} = \frac{\pi }{6} + k\pi \;\;\;\; \Leftrightarrow \frac{x}{2} = \frac{\pi }{{12}} + k\pi \;\;\; \Leftrightarrow x = \frac{\pi }{6} + k\pi \;\left( {k \in \mathbb{Z}} \right)\)

d) \(\cot \left( {2x - 1} \right) = \cot \frac{\pi }{5}\;\;\;\; \Leftrightarrow 2x - 1 = \frac{\pi }{5} + k\pi \;\;\;\; \Leftrightarrow 2x = \frac{\pi }{5} + 1 + k\pi \;\; \Leftrightarrow x = \frac{\pi }{{10}} + \frac{1}{2} + \frac{{k\pi }}{2}\;\;\left( {k \in \mathbb{Z}} \right)\)