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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) \(cos^2x-3sin^2x=0,19\Leftrightarrow1-sin^2x-3sin^2x=0,19\Leftrightarrow1-4sin^2x=0,19\Leftrightarrow4sin^2x=0,81\Leftrightarrow sin^2x=\dfrac{81}{400}\Leftrightarrow sinx=\dfrac{9}{20}\Leftrightarrow x=26^o44'\)
Có: \(1=\sin^2x+\cos^2x\ge2\sin x.\cos x\)\(\Leftrightarrow\)\(\sin x.\cos x\le\frac{1}{2}\)
\(M=\frac{1}{3\left(\frac{1}{\sin x}+\frac{1}{\cos x}\right)+\frac{2}{\sin x.\cos x}}\le\frac{1}{\frac{6}{\sqrt{\sin x.\cos x}}+\frac{2}{\sin x.\cos x}}\le\frac{1}{\frac{6}{\sqrt{\frac{1}{2}}}+\frac{2}{\frac{1}{2}}}=\frac{1}{6\sqrt{2}+4}\)
Dấu "=" xảy ra \(\Leftrightarrow\)\(\hept{\begin{cases}\frac{1}{\sin x}=\frac{1}{\cos x}\\\sin^2x+\cos^2x=1\end{cases}}\Leftrightarrow\sin x=\cos x=\frac{1}{\sqrt{2}}\)\(\Rightarrow\)\(x=45^0\)
\(\frac{2cos^2x-\left(cos^2x+sin^2x\right)}{cosx+sinx}=\frac{cos^2x-sin^2x}{cosx+sinx}=\frac{\left(cosx-sinx\right)\left(cosx+sinx\right)}{\left(cosx+sinx\right)}\)
\(=cosx-sinx\)
\(VT=\frac{2\cos^2x-1}{\cos x+\sin x}=\frac{2\cos^2x-\cos^2x-\sin^2x}{\cos x+\sin x}\)\(=\frac{\cos^2x-\sin^2x}{\cos x+\sin x}=\frac{\left(\cos x+\sin x\right)\left(\cos x-\sin x\right)}{\cos x+\sin x}\)
\(=\cos x-\sin x=VP\)
=> đpcm
Đề lỗi font. Bạn cần chỉnh sửa lại bằng công thức toán để được hỗ trợ tốt hơn.
a) \(cos^4x-sin^4x=\left(cos^2x+sin^2x\right)\left(cos^2x-sin^2x\right)=cos^2x-sin^2x\)
b) \(\frac{1}{1+tanx}+\frac{1}{1+cotx}=\frac{1}{1+tanx}+\frac{tanxcotx}{tanxcotx+cotx}=\frac{1}{1+tanx}+\frac{tanx}{tanx+1}\)
\(=\frac{1+tanx}{1+tanx}=1\)
c) Ta có: \(1+tan^2x=1+\frac{sin^2x}{cos^2x}=\frac{cos^2x+sin^2x}{cos^2x}=\frac{1}{cos^2x}\)
\(\Rightarrow\frac{1}{1+tan^2x}=cos^2x\)
Tương tự \(\frac{1}{1+tan^2y}=cos^2y\)
\(\Rightarrow cos^2x-cos^2y=\frac{1}{1+tan^2x}-\frac{1}{1+tan^2y}\)
\(cos^2x-cos^2y=\left(1-sin^2x\right)-\left(1-sin^2y\right)=sin^2y-sin^2x\)
d) \(\frac{1+sin^2x}{1-sin^2x}=\frac{cos^2x+sin^2x+sin^2x}{cos^2x+sin^2x-sin^2x}=\frac{cos^2x+2sin^2x}{cos^2x}=1+2\left(\frac{sinx}{cosx}\right)^2=1+2tan^2x\)