Question

Помогите 4,5,6
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С объяснением, пожалуйста.

Answer 1

$1)\; \; cosx=\frac{1}{\sqrt{10}}\\\\\frac{3\pi }{2}\leq x\leq 2\pi \; \; \to \; \; tgx\ \textless \ 0\\\\1+tg^2x=\frac{1}{cos^2x}\; \; \to \; \; tgx=\pm \sqrt{\frac{1}{cos^2x}-1}\\\\tgx\ \textless \ 0\; \; \to \; \; tgx=-\sqrt{\frac{1}{1/10}-1}=-\sqrt{9}=-3\\\\2)\; \; sinx=-\frac{5}{\sqrt{26}}\\\\\pi \leq x\leq \frac{3\pi }{2}\; \; \to \; \; tgx\ \textgreater \ 0\\\\tgx=\pm \sqrt{\frac{1}{cos^2x}-1}=\pm \sqrt{\frac{1}{1-sin^2x}-1}=\pm \sqrt{\frac{1-1+sin^2x}{1-sin^2x}}=\pm \sqrt{\frac{sin^2x}{1-sin^2x}}\\\\tgx\ \textgreater \ 0\; \; \to \; \; tgx=+\sqrt{\frac{25/26}{1-25/26}}=\sqrt{\frac{25}{26-25}}=5$

$3)\; \; sinx=-\frac{2\sqrt2}{3}\\\\\frac{3\pi }{2}\leq x\leq 2\pi \; \; \to \; \; cosx\ \textgreater \ 0\\\\sin^2x+cos^2x=1\; \; \to \; \; cos^2x=1-sin^2x\; ,\; cosx=\pm \sqrt{1-sin^2x}\\\\cosx\ \textgreater \ 0\; \; \to \; \; cosx=+\sqrt{1-(\frac{2\sqrt2}{3})^2}=\sqrt{1-\frac{8}{9}}=\sqrt{\frac{1}{9}}=\frac{1}{3}\\\\3cosx=3\cdot \frac{1}{3}=1$

$4)\; \; cosx= \frac{2\sqrt6}{5}\\\\\frac{3\pi}{2}\leq x\leq 2\pi \; \; \to \; \; sinx\ \textless \ 0\\\\sinx=-\sqrt{1-cos^2x}=-\sqrt{1-\frac{24}{25}}=-\frac{1}{5}\\\\5sinx=-5\cdot \frac{1}{5}=-1\\\\5)\; \; sinx=-0,2\\\\cos2x=cos^2x-sin^2x=(1-sin^2x)-sin^2x=1-2sin^2x=\\\\=1-2(-0,2)^2=1-2\cdot 0,04=0,92\\\\6)\; \; sin3x=0,6\\\\\frac{10sin6x}{3cos3x}=\frac{10\cdot 2sin3x\cdot cos3x}{3cos3x}=\frac{20}{3}\cdot sin3x=\frac{20}{3}\cdot 0,6=4$