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Đặt \(cosx-\frac{1}{cosx}=a\Rightarrow cos^2x+\frac{1}{cos^2x}=a^2+2\)

Pt trở thành:

\(a^2+2+a-\frac{7}{4}=0\)

\(\Leftrightarrow4a^2+4a+1=0\Leftrightarrow\left(2a+1\right)^2=0\)

\(\Rightarrow a=-\frac{1}{2}\Rightarrow cosx-\frac{1}{cosx}=-\frac{1}{2}\)

\(\Leftrightarrow2cos^2x+cosx-2=0\)

\(\Rightarrow\left[{}\begin{matrix}cosx=\frac{\sqrt{17}-1}{4}\\cosx=\frac{-\sqrt{17}-1}{4}< -1\left(l\right)\end{matrix}\right.\)

\(\Rightarrow x=\pm arccos\left(\frac{\sqrt{17}-1}{4}\right)+k2\pi\)

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\(\cos^21^o+\cos^289^o=\cos^21^o+\cos^2\left(90^o-1^o\right)=\cos^21^o+\sin^21^o=1\)

\(\cos^22^o+\cos^288^o=\cos^22^o+\cos^2\left(90^o-2^o\right)=\cos^22^o+\sin^22^o=1\)

.......

\(\cos^244^o+\cos^246^o=\cos^244^o+\cos^2\left(90^o-44^o\right)=\cos^244^o+\sin^244^o=1\)

\(\cos^245^o=\left(\frac{\sqrt{2}}{2}\right)^2=\frac{1}{2}\)

=> \(A=1.44+\frac{1}{2}-\frac{1}{2}=44\)

Lớp 9 không biết có học tới sin cos âm chưa nếu chưa thì lấy phần dương nha 

\(1+tan^2a=\frac{1}{cos^2a}\)    

\(1+\left(\frac{2}{3}\right)^2=\frac{1}{cos^2a}\)    

\(1+\frac{4}{9}=\frac{1}{cos^2a}\)    

\(\frac{13}{9}=\frac{1}{cos^2a}\)    

\(cos^2a=\frac{9}{13}\)   

\(cosa=\pm\sqrt{\frac{9}{13}}=\pm\frac{3\sqrt{13}}{13}\)    

\(sin^2a+cos^2a=1\)   

\(sin^2a+\frac{9}{13}=1\)    

\(sin^2a=\frac{4}{13}\)    

\(sina=\pm\sqrt{\frac{4}{13}}=\pm\frac{2\sqrt{13}}{13}\)   

tan dương nên sẽ có 2 TH 

TH1 sin và cos cùng dương 

\(\frac{sin^3a+3cos^3a}{27sin^3a-25cos^3a}\)   

\(=\frac{\left(\frac{2\sqrt{13}}{13}\right)^3+3\cdot\left(\frac{3\sqrt{13}}{13}\right)^3}{27\cdot\left(\frac{2\sqrt{13}}{13}\right)^3-25\cdot\left(\frac{3\sqrt{13}}{13}\right)^3}\)    

\(=-\frac{89}{459}\)   

TH2 sin và cos cùng âm 

\(\frac{sin^3a+3cos^3a}{27sin^3a-25cos^3a}\)   

\(=\frac{\left(\frac{-2\sqrt{13}}{13}\right)^3+\left(\frac{-3\sqrt{13}}{13}\right)^3}{27\cdot\left(\frac{-2\sqrt{13}}{13}\right)^3-25\cdot\left(\frac{-3\sqrt{13}}{13}\right)^3}\)

\(=-\frac{89}{459}\)

1.

Đặt \(\frac{x}{3}=t\) pt trở thành:

\(cos4t=sin^23t\Leftrightarrow2cos4t=1-cos6t\)

\(\Leftrightarrow cos6t+2cos4t-1=0\)

\(\Leftrightarrow4cos^32t-3cos2t+2\left(2cos^22t-1\right)-1=0\)

\(\Leftrightarrow4cos^32t+2cos^22t-3cos2t-3=0\)

\(\Leftrightarrow\left(cos2t-1\right)\left(4cos^22t+6cos2t+3\right)=0\)

\(\Leftrightarrow cos2t=1\Leftrightarrow cos\frac{2x}{3}=1\)

\(\Leftrightarrow\frac{2x}{3}=k2\pi\Leftrightarrow x=k3\pi\)

2.

\(\Leftrightarrow4cos^3x-3cosx-\left(1-2sin^2x\right)+9sinx-4=0\)

\(\Leftrightarrow cosx\left(4cos^2x-3\right)+2sin^2x+9sinx-5=0\)

\(\Leftrightarrow cosx\left(4\left(1-sin^2x\right)-3\right)+\left(2sinx-1\right)\left(sinx+5\right)=0\)

\(\Leftrightarrow cosx\left(1-4sin^2x\right)+\left(2sinx-1\right)\left(sinx+5\right)=0\)

\(\Leftrightarrow\left(cosx+2sinx.cosx\right)\left(1-2sinx\right)-\left(1-2sinx\right)\left(sinx+5\right)=0\)

\(\Leftrightarrow\left(1-2sinx\right)\left(cosx-sinx+2sinx.cosx-5\right)=0\)

\(\Leftrightarrow\left(1-2sinx\right)\left(\sqrt{2}cos\left(x+\frac{\pi}{4}\right)+sin2x-5\right)=0\)

\(\Leftrightarrow1-2sinx=0\) (do \(\sqrt{2}cos\left(x+\frac{\pi}{4}\right)\le\sqrt{2};sin2x\le1\) nên ngoặc sau luôn âm)

\(\Leftrightarrow sinx=\frac{1}{2}\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{6}+k2\pi\\x=\frac{5\pi}{6}+k2\pi\end{matrix}\right.\)

\(\frac{sin^3a+cos^3a}{sina+cosa}=\frac{\left(sina+cosa\right)\left(sin^2a+cos^2a-sina.cosa\right)}{sina+cosa}\)

\(=sin^2a+cos^2a-sina.cosa\)

\(=1-sina.cosa\)