Тригонометричекие уравненияЛюбое из первых двух заданий первой части (или оба...)

0\\\\cos\alpha =\sqrt{0,1}" alt="1)\quad \alpha \in (0;\frac{\pi}{2})\; ,\; \; cos \alpha =?\\\\ \frac{5-6ctg \alpha }{2-3ctg \alpha } =3\; \; \; \Rightarrow \; \; \; 5-6ctg \alpha =3(2-3ctg \alpha )\\\\5-6ctg \alpha =6-9ctg \alpha \\\\3ctg\alpha =1\\\\ctg \alpha =\frac{1}{3}\; \; \to \; \; \; tg \alpha \alpha =\frac{1}{ctg \alpha }=\frac{1}{1/3}=3\\\\1+tg^2 \alpha =\frac{1}{cos^2 \alpha }\; \; \to \; \; \; cos^2 \alpha =\frac{1}{1+tg^2 \alpha }=\frac{1}{1+3^2}=\frac{1}{10}=0,1\\\\ cos\alpha =\pm \sqrt{0,1}\\\\ \alpha \in (0;\frac{\pi}{2})\; \; \; \to \; \; \; cos\alpha >0\\\\cos\alpha =\sqrt{0,1}" align="absmiddle" class="latex-formula">

$2)\quad 15sin^2a-7sina=2\; ,\; \; a\in (\pi ,\frac{3\pi}{2})\\\\15sin^2a-7sina-1=0\; ,\; \; \; D=169\; ,\\\\sina=-\frac{1}{5}\; ,\; \; sina=\frac{2}{3}\\\\a\in (\pi ,\frac{3\pi}{2})\; \; \to \; \; sina\ \textless \ 0\; ,\; \; cosa\ \textless \ 0\\\\sina=-\frac{1}{5}\\\\cosa=-\sqrt{1-sin^2a}=-\sqrt{1-\frac{1}{25}}=-\sqrt\frac{24}{25}}=-\frac{2\sqrt{6}}{5}\\\\sin2a=2sina\cdot cosa=2\cdot (-\frac{1}{5})\cdot (-\frac{2\sqrt6}{5})=\frac{4\sqrt6}{25}$

$cos2a=cos^2a-sin^2a=(1-sin^2a)-sin^2a=1-2sin^2a=\\\\=1-2\cdot (-\frac{1}{5})^2=1-\frac{2}{25}=\frac{23}{25}$