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Bài 1:
ĐK : sinx cosx > 0
Khi đó phương trình trở thành
sinx+cosx=\(2\sqrt{\sin x\cos x}\)
ĐK sinx + cosx >0 → sinx>0 ; cosx>0
Khi đó \(2\sqrt{\sin x\cos x}\Leftrightarrow2\sin x=1\Leftrightarrow x=\frac{\pi}{4}+k\pi\)
Vậy ...
Bài 2:
ĐK : \(\sin\left(3x+\frac{\pi}{4}\right)\ge0\)
Khi đó phương trình đã cho tương đương với phương trình \(\sin2x=\frac{1}{2}\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\frac{\pi}{12}+k\pi\\x=\frac{5\pi}{12}+k\pi\end{matrix}\right.\)
Trong khoảng từ \(\left(-\pi,\pi\right)\) ta nhận được các giá trị :
\(x=\frac{\pi}{12}\) (TMĐK)
\(x=-\frac{11\pi}{12}\) (KTMĐK)
\(x=\frac{5\pi}{12}\) (KTMĐK)
\(x=-\frac{7\pi}{12}\) (TMĐK)
Vậy ta có 2 nghiệm thõa mãn \(x=\frac{\pi}{12}\) và \(x=-\frac{7\pi}{12}\)
3.
\(\Leftrightarrow\dfrac{\sqrt{3}}{2}sinx-\dfrac{1}{2}cosx=cos3x\)
\(\Leftrightarrow sin\left(x-\dfrac{\pi}{6}\right)=sin\left(\dfrac{\pi}{2}-3x\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{\pi}{6}=\dfrac{\pi}{2}-3x+k2\pi\\x-\dfrac{\pi}{6}=\dfrac{\pi}{2}+3x+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{6}+\dfrac{k\pi}{2}\\x=-\dfrac{\pi}{3}+k\pi\end{matrix}\right.\)
\(\Leftrightarrow\cos4x+\cos2x+\sqrt{3}\left(1+\sin2x\right)=\sqrt{3}\left(1+\cos\left(4x+\frac{\pi}{2}\right)\right)\)
\(\Leftrightarrow\cos4x+\sqrt{3}\sin4x+\sqrt{3}\sin2x=0\)
\(\Leftrightarrow\sin\left(4x+\frac{\pi}{6}\right)+\sin\left(2x+\frac{\pi}{6}\right)=0\)
\(\Leftrightarrow2\sin\left(3x+\frac{\pi}{6}\right)\cos x=0\)
\(\Leftrightarrow\begin{cases}x=-\frac{\pi}{18}+k\frac{\pi}{3}\\x=\frac{\pi}{2}+k\pi\end{cases}\)
Vậy phương trình có 2 nghiệm \(x=-\frac{\pi}{18}+k\frac{\pi}{3}\) và \(x=\frac{\pi}{2}+k\pi\)
\(4sin\left(x+\dfrac{\pi}{3}\right).cos\left(x-\dfrac{\pi}{6}\right)=m^2+\sqrt[]{3}sin2x-cos2x\)
\(\Leftrightarrow4.\left(-\dfrac{1}{2}\right)\left[sin\left(x+\dfrac{\pi}{3}+x-\dfrac{\pi}{6}\right)+sin\left(x+\dfrac{\pi}{3}-x+\dfrac{\pi}{6}\right)\right]=m^2+2.\left[\dfrac{\sqrt[]{3}}{2}.sin2x-\dfrac{1}{2}.cos2x\right]\)
\(\Leftrightarrow2\left[sin\left(2x+\dfrac{\pi}{6}\right)+sin\left(2x-\dfrac{\pi}{6}\right)\right]=m^2+2\)
\(\Leftrightarrow2.2sin2x.cos\dfrac{\pi}{6}=m^2+2\)
\(\Leftrightarrow2.2sin2x.\dfrac{\sqrt[]{3}}{2}=m^2+2\)
\(\Leftrightarrow2\sqrt[]{3}sin2x.=m^2+2\)
\(\Leftrightarrow sin2x.=\dfrac{m^2+2}{2\sqrt[]{3}}\)
Phương trình có nghiệm khi và chỉ khi
\(\left|\dfrac{m^2+2}{2\sqrt[]{3}}\right|\le1\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{m^2+2}{2\sqrt[]{3}}\ge-1\\\dfrac{m^2+2}{2\sqrt[]{3}}\le1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}m^2\ge-2\left(1+\sqrt[]{3}\right)\left(luôn.đúng\right)\\m^2\le2\left(1-\sqrt[]{3}\right)\end{matrix}\right.\)
\(\Leftrightarrow-\sqrt[]{2\left(1-\sqrt[]{3}\right)}\le m\le\sqrt[]{2\left(1-\sqrt[]{3}\right)}\)
\(cos\left(\dfrac{\pi}{6}-2x\right)=cos\left(\dfrac{\pi}{2}-x\right)\)
\(\Rightarrow\left[{}\begin{matrix}\dfrac{\pi}{6}-2x=\dfrac{\pi}{2}-x+k2\pi\\\dfrac{\pi}{6}-2x=x-\dfrac{\pi}{2}+k2\pi\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{\pi}{3}+k2\pi\\x=\dfrac{2\pi}{9}+\dfrac{k2\pi}{3}\end{matrix}\right.\)
\(\Rightarrow x=\left\{\dfrac{8\pi}{9};\dfrac{14\pi}{9};\dfrac{5\pi}{3}\right\}\) có 3 nghiệm
a) \(\cos \left( {3x - \frac{\pi }{4}} \right) = - \frac{{\sqrt 2 }}{2}\;\;\;\; \Leftrightarrow \cos \left( {3x - \frac{\pi }{4}} \right) = \cos \frac{{3\pi }}{4}\;\;\; \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{3x - \frac{\pi }{4} = \frac{{3\pi }}{4} + k2\pi }\\{3x - \frac{\pi }{4} = - \frac{{3\pi }}{4} + k2\pi }\end{array}} \right.\;\;\;\; \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{3x = \pi + k2\pi }\\{3x = - \frac{\pi }{2} + k2\pi }\end{array}} \right.\)
\( \Leftrightarrow \;\left[ {\begin{array}{*{20}{c}}{x = \frac{\pi }{3} + \frac{{k2\pi }}{3}}\\{x = - \frac{\pi }{6} + \frac{{k2\pi }}{3}}\end{array}} \right.\;\;\left( {k \in \mathbb{Z}} \right)\)
b) \(2{\sin ^2}x - 1 + \cos 3x = 0\;\;\;\;\; \Leftrightarrow \cos 2x + \cos 3x = 0\;\; \Leftrightarrow 2\cos \frac{{5x}}{2}\cos \frac{x}{2} = 0\;\; \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{\cos \frac{{5x}}{2} = 0}\\{\cos \frac{x}{2} = 0}\end{array}} \right.\)
\( \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{\frac{{5x}}{2} = \frac{\pi }{2} + k\pi }\\{\frac{{5x}}{2} = - \frac{\pi }{2} + k\pi }\\{\frac{x}{2} = \frac{\pi }{2} + k\pi }\\{\frac{x}{2} = - \frac{\pi }{2} + k\pi }\end{array}} \right.\;\;\;\;\;\;\; \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{x = \frac{\pi }{5} + \frac{{k2\pi }}{5}}\\{x = - \frac{\pi }{5} + \frac{{k2\pi }}{5}}\\{x = \pi + k2\pi }\\{x = - \pi + k2\pi }\end{array}} \right.\;\;\;\left( {k \in \mathbb{Z}} \right)\)
c) \(\tan \left( {2x + \frac{\pi }{5}} \right) = \tan \left( {x - \frac{\pi }{6}} \right)\;\; \Leftrightarrow 2x + \frac{\pi }{5} = x - \frac{\pi }{6} + k\pi \;\;\; \Leftrightarrow x = - \frac{{11\pi }}{{30}} + k\pi \;\;\left( {k \in \mathbb{Z}} \right)\)
a/ Hmm, bạn có nhầm lẫn chỗ nào ko nhỉ, nghiệm của pt này xấu khủng khiếp
b/ \(\Leftrightarrow sin\frac{5x}{2}-cos\frac{5x}{2}-sin\frac{x}{2}-cos\frac{x}{2}=cos\frac{3x}{2}\)
\(\Leftrightarrow2cos\frac{3x}{2}.sinx-2cos\frac{3x}{2}cosx=cos\frac{3x}{2}\)
\(\Leftrightarrow cos\frac{3x}{2}\left(2sinx-2cosx-1\right)=0\)
\(\Leftrightarrow cos\frac{3x}{2}\left(\sqrt{2}sin\left(x-\frac{\pi}{4}\right)-1\right)=0\)
c/ Do \(cosx\ne0\), chia 2 vế cho cosx ta được:
\(3\sqrt{tanx+1}\left(tanx+2\right)=5\left(tanx+3\right)\)
Đặt \(\sqrt{tanx+1}=t\ge0\)
\(\Leftrightarrow3t\left(t^2+1\right)=5\left(t^2+2\right)\)
\(\Leftrightarrow3t^3-5t^2+3t-10=0\)
\(\Leftrightarrow\left(t-2\right)\left(3t^2+t+5\right)=0\)
d/ \(\Leftrightarrow\sqrt{2}\left(\frac{1}{2}sinx+\frac{\sqrt{3}}{2}cosx\right)=\frac{\sqrt{3}}{2}cos2x-\frac{1}{2}sin2x\)
\(\Leftrightarrow\sqrt{2}sin\left(x+\frac{\pi}{3}\right)=-sin\left(2x-\frac{\pi}{3}\right)\)
Đặt \(x+\frac{\pi}{3}=a\Rightarrow2x=2a-\frac{2\pi}{3}\Rightarrow2x-\frac{\pi}{3}=2a-\pi\)
\(\sqrt{2}sina=-sin\left(2a-\pi\right)=sin2a=2sina.cosa\)
\(\Leftrightarrow\sqrt{2}sina\left(\sqrt{2}cosa-1\right)=0\)
\(cos^2x-sin2x=\sqrt{2}+sin^2x\)
\(\Leftrightarrow cos^2x-sin^2x-sin2x=\sqrt{2}\)
\(\Leftrightarrow cos2x-sin2x=\sqrt{2}\)
\(\Leftrightarrow\sqrt{2}cos\left(2x+\frac{\pi}{4}\right)=\sqrt{2}\)
\(\Leftrightarrow cos\left(2x+\frac{\pi}{4}\right)=1\)
\(\Leftrightarrow x=-\frac{\pi}{8}+k\pi\)
\(0< -\frac{\pi}{8}+k\pi< 3\pi\Rightarrow\frac{1}{8}< k< \frac{25}{8}\Rightarrow k=\left\{1;2;3\right\}\)
Có 3 nghiệm