Question

4×sin2a=(1-cos4a)×(ctga+tga)​

Answer 1

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

$left { {{4sin2a=frac{4sin^22a}{sin2a} =4sin2a} atop {{left[begin{array}{ccc}sinaneq 0\cosaneq 0\sin2aneq 0end{array}}} right.$

${{left[begin{array}{ccc}sinaneq 0\cosaneq 0\2*cosa*sinaneq 0end{array}}}$

Ответ: x≠{pi*n/2}, n∈Z.