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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) \(\cot \left( {2x - 1} \right) = \cot \frac{\pi }{5}\;\;\;\; \Leftrightarrow 2x - 1 = \frac{\pi }{5} + k\pi \;\;\;\; \Leftrightarrow 2x = \frac{\pi }{5} + 1 + k\pi \;\; \Leftrightarrow x = \frac{\pi }{{10}} + \frac{1}{2} + \frac{{k\pi }}{2}\;\;\left( {k \in \mathbb{Z}} \right)\)
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.
\(\left(sin^2x+cos^2x\right)^3-3sin^2x.cos^2x\left(sin^2x+cos^2x\right)=\dfrac{7}{16}\)
\(\Leftrightarrow1-\dfrac{3}{4}\left(2sinx.cosx\right)^2=\dfrac{7}{16}\)
\(\Leftrightarrow16-12.sin^22x=7\)
\(\Leftrightarrow3-4sin^22x=0\)
\(\Leftrightarrow3-2\left(1-cos4x\right)=0\)
\(\Leftrightarrow cos4x=-\dfrac{1}{2}\)
\(\Leftrightarrow4x=\pm\dfrac{2\pi}{3}+k2\pi\)
\(\Leftrightarrow x=\pm\dfrac{\pi}{6}+\dfrac{k\pi}{2}\)
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
b)
(sin2x + cos2x)cosx + 2cos2x - sinx = 0
⇔ cos2x (cosx + 2) + sinx (2cos2 x – 1) = 0
⇔ cos2x (cosx + 2) + sinx.cos2x = 0
⇔ cos2x (cosx + sinx + 2) = 0
⇔ cos2x = 0
⇔ 2x = 
+ kπ ⇔ x = 
+ k
(k ∈ 
)
c)
Đáp án:
x=π6π6+ k2ππ
và x= 5π65π6+k2ππ (k∈Z)
Lời giải:
sin2x-cos2x+3sinx-cosx-1=0
⇔ 2sinxcosx-(1-2sin²x) +3sinx-cosx-1=0
⇔ 2sin²x+2sinxcosx+3sinx-cosx-2=0
⇔ (2sin²x+3sinx-2)+ cosx(2sinx-1)=0
⇔ (2sinx-1)(sinx+2)+cosx(2sinx-1)=0
⇔ (2sinx-1)(sinx+cosx+2)=0
⇔ sinx=1212
⇔ x=π6π6+ k2ππ
hoặc x= 5π65π6+k2ππ (k∈Z)
(sinx+cosx+2)=0 (vô nghiệm do sinx+cosx+2=√22sin(x+π4π4)+2>0)
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 \(sinx+cosx=t\Rightarrow\left\{{}\begin{matrix}\left|t\right|\le\sqrt{2}\\sinx.cosx=\frac{t^2-1}{2}\end{matrix}\right.\)
\(\Rightarrow t=\frac{t^2-1}{2}\Rightarrow t^2-2t-1=0\Rightarrow\left[{}\begin{matrix}t=1+\sqrt{2}\left(l\right)\\t=1-\sqrt{2}\end{matrix}\right.\)
\(\Rightarrow\sqrt{2}sin\left(x+\frac{\pi}{4}\right)=1-\sqrt{2}\Rightarrow sin\left(x+\frac{\pi}{4}\right)=\frac{1-\sqrt{2}}{\sqrt{2}}\Rightarrow x=...\)
Ta có:
a) \(\sin \left( {x + 2\pi } \right) = \sin x\) với mọi \(x\; \in \;\mathbb{R}\)
b) \(\cos \left( {x + 2\pi } \right) = \cos x\) với mọi \(x\; \in \;\mathbb{R}\)
c) \(\tan \left( {x + \pi } \right) = \tan x\) với mọi \(x \ne \frac{\pi }{2} + k\pi ,\;k\; \in \;\mathbb{Z}\)
d) \(\cot \left( {x + \pi } \right) = \cot x\) với mọi \(x \ne \frac{\pi }{2} + k\pi ,\;k\; \in \;\mathbb{Z}\)
\(PT\Leftrightarrow2+\frac{1}{sinxcosx}-cotx=-2-sinx-cosxcotx-tanx\)
\(\Leftrightarrow\frac{1}{sinxcosx}-\frac{cosx}{sinx}=-2sinx-\frac{sinx}{cosx}\)
\(\Leftrightarrow1-cos^2x+2sin^2xcosx+sin^2x=0\)
\(\Leftrightarrow2sin^2x+2sin^2xcosx=0\)
\(\Leftrightarrow2sin^2x\left(1+cosx\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sin^2x=0\\cosx=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}cosx=1\\cosx=-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2k\pi\\x=\pi+2k\pi\end{matrix}\right.\Leftrightarrow x=k\pi\)
Mình sửa lại câu trả lời
ĐK:\(\left\{{}\begin{matrix}sinx\ne0\\cosx\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne k\pi\\x\ne\frac{\pi}{2}+k\pi\end{matrix}\right.\)
\(PT\Leftrightarrow2sin^2x\left(1+cosx\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=0\left(loai\right)\\cosx=-1\end{matrix}\right.\)
\(\Leftrightarrow x=\pi+2k\pi\left(loai\right)\)
Vậy phương trình vô nghiệm