Question

4×sin2a=(1-cos4a)×(ctga+tga)​

Answer 1

$4sin2a=(1-(1-2sin^22a))*(\frac{cos^2a+sin^2a}{sina*cosa} );\\\left \{ {{4sin2a=\frac{2sin^22a*1}{\frac{sin2a}{2} }} \atop {{\left[\begin{array}{ccc}sina\neq 0\\cosa\neq 0\\\end{array} }}\right.$

$\left \{ {{4sin2a=\frac{4sin^22a}{sin2a} =4sin2a} \atop {{\left[\begin{array}{ccc}sina\neq 0\\cosa\neq 0\\sin2a\neq 0\end{array}}} \right.$

${{\left[\begin{array}{ccc}sina\neq 0\\cosa\neq 0\\2*cosa*sina\neq 0\end{array}}}$

Ответ: x≠{pi*n/2}, n∈Z.