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NV
Nguyễn Việt Lâm
Giáo viên
25 tháng 9 2019
ĐKXĐ: \(cosx\ne\frac{1}{2}\Rightarrow x\ne\pm\frac{\pi}{3}+k2\pi\)
\(cos2x+\sqrt{3}\left(1+sinx\right)=\frac{2cosx-1+4sinx.cosx-2sinx}{2cosx-1}\)
\(\Leftrightarrow cos2x+\sqrt{3}\left(1+sinx\right)=\frac{2cosx-1+2sinx\left(2cosx-1\right)}{2cosx-1}\)
\(\Leftrightarrow cos2x+\sqrt{3}+\sqrt{3}sinx=2sinx+1\)
\(\Leftrightarrow1-2sin^2x+\sqrt{3}\left(1+sinx\right)=2sinx+1\)
\(\Leftrightarrow2sin^2x+2sinx-\sqrt{3}\left(1+sinx\right)=0\)
\(\Leftrightarrow\left(2sinx-\sqrt{3}\right)\left(1+sinx\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}sinx=-1\\sinx=\frac{\sqrt{3}}{2}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=-\frac{\pi}{2}+k2\pi\\x=\frac{\pi}{3}+k2\pi\left(ktm\right)\\x=\frac{2\pi}{3}+k2\pi\end{matrix}\right.\)
17 tháng 5 2021
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)\(\left(2sin-\sqrt{2}\right)\left(4cos-5\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}2sin-\sqrt{2}=0\\4cos-5=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}2sin=\sqrt{2}\\4cos=5\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}sin=\dfrac{\sqrt{2}}{2}\\cos=\dfrac{5}{4}\left(loai\right)\end{matrix}\right.\)\(\Rightarrow x=45^o\)
17 tháng 5 2021
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) \(cos^2x-3sin^2x=0,19\Leftrightarrow1-sin^2x-3sin^2x=0,19\Leftrightarrow1-4sin^2x=0,19\Leftrightarrow4sin^2x=0,81\Leftrightarrow sin^2x=\dfrac{81}{400}\Leftrightarrow sinx=\dfrac{9}{20}\Leftrightarrow x=26^o44'\)
8 tháng 8 2022
1: \(=\dfrac{cotx+1+tanx+1}{\left(tanx+1\right)\left(cotx+1\right)}\)
\(=\dfrac{\dfrac{1}{cotx}+cotx+2}{2+tanx+cotx}\)
\(=1\)
2: \(VT=\dfrac{cos^2x+cosxsinx+sin^2x-sinx\cdot cosx}{sin^2x-cos^2x}\)
\(=\dfrac{1}{sin^2x-cos^2x}\)
\(VP=\dfrac{1+cot^2x}{1-cot^2x}=\left(1+\dfrac{cos^2x}{sin^2x}\right):\left(1-\dfrac{cos^2x}{sin^2x}\right)\)
\(=\dfrac{1}{sin^2x}:\dfrac{sin^2x-cos^2x}{sin^2x}=\dfrac{1}{sin^2x-cos^2x}\)
=>VT=VP
24 tháng 9 2020
\(B\sqrt{2}=\sqrt{6+2\sqrt{5}}-\sqrt{6-2\sqrt{5}}-2\)\(=\sqrt{\left(\sqrt{5}+1\right)^2}-\sqrt{\left(\sqrt{5}-1\right)^2}-2\)\(=\left|\sqrt{5}+1\right|-\left|\sqrt{5}-1\right|-2=\sqrt{5}+1-\sqrt{5}+1-2=0\Rightarrow B=0\)
\(C=\left(1+\frac{\sin^2a}{\cos^2a}\right)\left(1-\sin^2a\right)+\left(1+\frac{\cos^2a}{\sin^2a}\right)\left(1-\cos^2a\right)\)
\(=\left(1+\frac{\sin^2a}{\cos^2a}\right)\left(\cos^2a\right)+\left(1+\frac{\cos^2a}{\sin^2a}\right)\left(\sin^2a\right)\)
\(=\frac{\sin^2a+\cos^2a}{\cos^2a}.\cos^2a+\frac{\cos^2a+\sin^2a}{\sin^2a}.\sin^2a\)
\(=\frac{1}{\cos^2a}.\cos^2a+\frac{1}{\sin^2a}\sin^2a=2\)
24 tháng 9 2020
B
Bạn dùng theo công thức này
\(\sqrt{m+n\sqrt{p}};\sqrt{m-n\sqrt{p}}\)
Dùng pt bậc 2
\(a=1;b=-m;c=\frac{\left(n\sqrt{p}\right)^2}{4}\)
Nghiệm x1 ; x2
\(\sqrt{\left(\sqrt{x1}+\sqrt{x2}\right)^2};\sqrt{\left(\sqrt{x1}-\sqrt{x2}\right)^2}\)
\(B=\sqrt{\left(\sqrt{\frac{5}{2}}+\sqrt{\frac{1}{2}}\right)^2}-\sqrt{\left(\sqrt{\frac{5}{2}}-\sqrt{\frac{1}{2}}\right)^2}-\sqrt{2}\)
\(=|\sqrt{\frac{5}{2}}+\sqrt{\frac{1}{2}}|-|\sqrt{\frac{5}{2}}-\sqrt{\frac{1}{2}}|-\sqrt{2}\)
\(=\sqrt{\frac{5}{2}}+\sqrt{\frac{1}{2}}-\left(\sqrt{\frac{5}{2}}-\sqrt{\frac{1}{2}}\right)-\sqrt{2}\)
\(=2\cdot\sqrt{\frac{1}{2}}-\sqrt{2}\)
\(=\sqrt{2}-\sqrt{2}=0\)
C.
\(=\frac{1}{cos^2a}\cdot cos^2a+\frac{1}{sin^2a}\cdot sin^2a\)
\(=1+1=2\)
NH
1
4 tháng 9 2022
\(\Leftrightarrow\dfrac{1-cos2x}{2}-\left(1+\sqrt{3}\right)\cdot\dfrac{1}{2}sin2x+\sqrt{3}\cdot\dfrac{1+cos2x}{2}=0\)
\(\Leftrightarrow\dfrac{1}{2}-\dfrac{1}{2}cos2x-\left(\dfrac{1}{2}+\dfrac{\sqrt{3}}{2}\right)\cdot sin2x+\dfrac{\sqrt{3}}{2}+\dfrac{\sqrt{3}}{2}cos2x=0\)
\(\Leftrightarrow cos2x\left(\dfrac{\sqrt{3}}{2}-\dfrac{1}{2}\right)-\left(\dfrac{1}{2}+\dfrac{\sqrt{3}}{2}\right)\cdot sin2x=\dfrac{-\sqrt{3}-1}{2}\)
\(\Leftrightarrow sin2x\cdot\dfrac{-\sqrt{3}-1}{2}+cos2x\cdot\dfrac{\sqrt{3}-1}{2}=\dfrac{-\sqrt{3}-1}{2}\)
\(\Leftrightarrow sin2x\left(-\sqrt{3}-1\right)+cos2x\left(\sqrt{3}-1\right)=-\sqrt{3}-1\)
\(\Leftrightarrow sin2x\cdot\dfrac{-\sqrt{3}-1}{8}+cos2x\cdot\dfrac{\sqrt{3}-1}{8}=\dfrac{-\sqrt{3}-1}{8}\)
\(\Leftrightarrow sin\left(2x+a\right)=cosa=sin\left(\dfrac{pi}{2}-a\right)\)(với \(cosa=\dfrac{-\sqrt{3}-1}{8}\))
\(\Leftrightarrow\left[{}\begin{matrix}2x+a=\dfrac{pi}{2}-a+k2pi\\2x+a=pi-\dfrac{pi}{2}+a+k2pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=-2a+\dfrac{pi}{2}+k2pi\\2x=\dfrac{pi}{2}+k2pi\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-a+\dfrac{pi}{4}+kpi\\x=\dfrac{pi}{4}+kpi\end{matrix}\right.\)
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