Нужно решить уравнение: arcsin 5x = arcctg 6x

$arcsin5x=arcctg6x\\\\sin(arcsin5x)=sin(arcctg6x)\\\\\star \; \; sin(arcsin \alpha )= \alpha \; ,\; \; -1\leq \alpha \leq 1\; \; \star \\\\\star \; \; 1+ctg^2\alpha \; =\frac{1}{sin^2 \alpha }\; ,\; sin^2\alpha =\frac{1}{1+ctg^2\alpha }\; ,\; \; ctg(arcctg \alpha )= \alpha \; \; \star \\\\sin \alpha =sin(\underbrace {arcctg6x}_{\alpha })=\frac{1}{\sqrt{1+ctg^2 \alpha }}=\frac{1}{\sqrt{1+ctg^2(arcctg6x)}} = \frac{1}{\sqrt{1+(6x)^2}} \\\\sin(arcctg6x)= \frac{1}{1+36x^2} \\\\5x= \frac{1}{\sqrt{1+36x^2}}$

$\sqrt{36x^2+1} = \frac{1}{5x} \\\\36x^2+1=\frac{1}{25x^2}\\\\36\cdot 25x^4+25x^2-1=0\\\\D=25^2+4\cdot 36\cdot 25=4225=65^2\\\\(x^2)_1= \frac{-25-65}{2\cdot 36\cdot 25} = \frac{-90}{2\cdot 9\cdot 4\cdot 5\cdot 5 }= \frac{-1}{20} \ \textless \ 0\; \; ne\; podxodit,\; t.k.\; x^2 \geq 0\\\\(x^2)_2= \frac{-25+65}{2\cdot 36\cdot 25} = \frac{40}{40\cdot 45} =\frac{1}{45}\ \textgreater \ 0\\\\x^2=\frac{1}{45}\\\\x=\pm \frac{1}{\sqrt{45}}=\pm \frac{1}{3\sqrt5}=\pm \frac{\sqrt5}{15}$

$Otvet:\; \; x=- \frac{\sqrt5}{15}\; ,\; \frac{\sqrt5}{15}\; .$