a/ Nhận thấy \(cosx=0\) ko phải nghiệm, chia 2 vế cho \(cos^3x\)
\(\Leftrightarrow2tan^3x=\frac{1}{cos^2x}\)
\(\Leftrightarrow2tan^3x=1+tan^2x\)
\(\Leftrightarrow2tan^3x-tan^2x-1=0\)
\(\Leftrightarrow\left(tanx-1\right)\left(2tan^2x+tanx+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}tanx=1\\2tan^2x+tanx+1=0\left(vn\right)\end{matrix}\right.\)
\(\Rightarrow x=\frac{\pi}{4}+k\pi\)
b/
\(\Leftrightarrow sin^3x-sinx+cos^3x+cosx=0\)
\(\Leftrightarrow-sinx\left(1-sin^2x\right)+cos^3x+cosx=0\)
\(\Leftrightarrow-sinx.cos^2x+cos^3x+cosx=0\)
\(\Leftrightarrow cosx\left(cos^2x-sinx.cosx+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cosx=0\Rightarrow x=\frac{\pi}{2}+k\pi\\cos^2x-sinx.cosx+1=0\left(1\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow\frac{1}{2}+\frac{1}{2}cos2x-\frac{1}{2}sin2x+1=0\)
\(\Leftrightarrow cos2x-sin2x=-3\)
\(\Leftrightarrow sin\left(2x-\frac{\pi}{4}\right)=\frac{3}{\sqrt{2}}>1\left(vn\right)\)
c/
Nhận thấy \(cosx=0\) ko phải nghiệm, chia 2 vế cho \(cos^3x\)
\(4+2tan^3x-3tanx.\frac{1}{cos^2x}=0\)
\(\Leftrightarrow2tan^3x-3tanx\left(1+tan^2x\right)+4=0\)
\(\Leftrightarrow-tan^3x-3tanx+4=0\)
\(\Leftrightarrow\left(1-tanx\right)\left(tan^2x+tanx+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}tanx=1\\tan^2x+tanx+4=0\left(vn\right)\end{matrix}\right.\)
\(\Rightarrow x=\frac{\pi}{4}+k\pi\)
d/
\(\Leftrightarrow2cos^3x+2sinx-6sin^2x.cosx=0\)
Nhận thấy \(cosx=0\) ko phải nghiệm, chia 2 vế cho \(cos^3x\)
\(2+2tanx.\frac{1}{cos^2x}-6tan^2x=0\)
\(\Leftrightarrow1+tanx\left(1+tan^2x\right)-3tan^2x=0\)
\(\Leftrightarrow tan^3x-3tan^2x+tanx+1=0\)
\(\Leftrightarrow\left(tanx-1\right)\left(tan^2x-2tanx-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}tanx=1\\tan^2x-2tanx-1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}tanx=1\\tanx=1-\sqrt{2}\\tanx=1+\sqrt{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{4}+k\pi\\x=\frac{3\pi}{8}+k\pi\\x=-\frac{\pi}{8}+k\pi\end{matrix}\right.\)
