Question

1)6cosx+5sinx^2-6=0
2) 2log1/9 2-3x/x> -1

Answer 1

$\displaystyle \\6\cos{x}+5\sin^2{x}-6=0\\6\cos{x}+5(1-\cos^2{x})-6=0\\6\cos{x}-1-5\cos^2(x)=0\\\\t=\cos{x}\\\\5t^2-6t+1=0\\t_1 = 1, t_2 = \frac{1}{5}\\\\\cos{x}=1, cos{x}=\frac{1}{5} \\\begin{cases}x=2\pi n\\x = \pm \arccos{\frac{1}{5}}+2\pi n\end{cases}, n \in \mathbb Z$

$\displaystyle \\2\log_{1/9}(\frac{2-3x}{x}) -1 \\ \\D(f) = (0, \frac{2}{3})\\ \\\log_{1/9}(\frac{2-3x}{x})-\frac{1}{2}\\ \\\frac{2-3x}{x} < {(1/9)}^{-(1/2)}\\ \\\frac{2-3x}{x} < 3\\ \\\frac{2-6x}{x} < 0\\ \\\frac{1-3x}{x} < 0\\ \\x \in (-\infty, 0) \cup(\frac{1}{3}, +\infty)\\$

с учётом D(f): $\displaystyle x \in (\frac{1}{3}, \frac{2}{3})$