$c)tan(\pi /4-2x)=\frac{1-tan2x}{tan2x+1}tan(π/4+2x)=1+tan2x1−tan2x$pt$⇔−1=4cos22xtanx−cotx-1=\frac{4(cosx^2-sinx^2).cos2x}{\frac{sin^2x-cos^2x}{sinxcosx}}−1=−4.sinx.cosx.cos2xsin4x=1\Leftrightarrow 4x=\frac{\pi }{2}+k2\pi \Leftrightarrow x=\frac{\pi }{8}+\frac{k\pi }{2}$
c) ĐK $c
os(2x\pm\pi /4)
\neq 0; sin2x\neq 0$$tan(\pi /4-2x)=\frac{1-tan2x}{
1+tan2x}
tan(π/4+2x)=1+tan2x1−tan2x$pt$⇔−1=4cos22xtanx−cotx-1=\frac{4(cosx^2-sinx^2).cos2x}{\frac{sin^2x-cos^2x}{sinxcosx}}
−1=−4.sinx.cosx.cos2xsin4x=1\Leftrightarrow 4x=\frac{\pi }{2}+k2\pi \Leftrightarrow x=\frac{\pi }{8}+\frac{k\pi }{2}$