Номер 266(2,4) и номер 265(1)

Question 1

Question 2

Решите задачу:

$265(1).\; \; sin^26x+8sin^23x=0\\\\\star \; \; sin^2 \alpha = \frac{1-cos2\alpha }{2} \; \; \star \\\\ \frac{1-cos12x}{2}+8\cdot \frac{1-cos6x}{2}=0\\\\1-cos12x+8-8cos6x=0\\\\cos12x+8cos6x-9=0\\\\\star cos2 \alpha =cos^2 \alpha -sin^2\alpha =cos^2\alpha -(1-cos^2 \alpha )=2cos^2\alpha -1\; \star \\\\2cos^26x-1+8cos6x-9=0\; |:2\\\\cos^26x+4cos6x-5=0\\\\t=cos6x\; ,\; \; \; t^2+4t-5=0\; ,\; \; t_1=-5\; ,\; \; t_2=1\; (teorema\; Vieta)\\\\a)\; \; cos6x=-5\ \textgreater \ 1\; \; net\; \; reshenij,\; t.k.\; \; -1 \leq cos6x \leq 1$

$b)\; \; cos6x=1\; ,\; \; 6x=2\pi n\; ,\; \; \underline {x=\frac{\pi n}{3}\; ,\; n\in Z }$

$266(2).\; \; cos^2\frac{x}{2}+cos^2\frac{3x}{2}=sin^22x+sin^24x\\\\\star \; \; cos^2 \alpha =\frac{1+cos2 \alpha }{2}\; \; ,\; \; sin^2 \alpha =\frac{1-cos2 \alpha }{2}\; \; \star \\\\ \frac{1+cosx}{2}+\frac{1+cos3x}{2}=\frac{1-cos4x}{2}+\frac{1-cos8x}{2}\\\\cosx+cos3x=-(cos4x+cos8x)\\\\2cos2x\cdot cosx=-2cos6x\cdot cos2x\\\\2cos2x\cdot (cosx+cos6x)=0\\\\cos2x\cdot 2cos\frac{7x}{2}\cdot cos\frac{5x}{2}=0\\\\a)\; \; cos2x=0\; ,\; 2x=2\pi n,\; \; \underline {x=\pi n,\; n\in Z}\\\\b)\; \; cos\frac{7x}{2}=0\; ,\; \; \frac{7x}{2}=2\pi k\; ,\; \; \underline {x=\frac{4\pi k}{7}\; ,\; k\in Z}$

$c)\; \; cos \frac{5x}{2}=0\; ,\; \; \frac{5x}{2}=2\pi m\; ,\; \; \underline {x= \frac{4\pi m}{5}\; ,\; n\in Z}$

$266(4).\; \; 8sin^2x+6cos^2x=13sin2x\\\\2sin^2x+(6sin^2x+6cos^2x)=13sin2x\\\\2sin^2x+6=13sin2x\\\\(1-cos2x)+6=13sin2x\\\\13sin2x+cos2x=7\; |:\sqrt{170}\\\\ \underbrace {\frac{13}{\sqrt{170}}}_{sin \alpha }\cdot sin2x+\underbrace {\frac{1}{\sqrt{170}}}_{cos \alpha }\cdot cos2x=\frac{7}{\sqrt{170}} \\\\ cos(2x- \alpha )=\frac{7}{\sqrt{170}}\; ,\; \alpha =arctg13\\\\2x- \alpha =\pm arccos\frac{7}{\sqrt{170}}+2\pi n,\; n\in Z\\\\\underline {x=\frac{1}{2}\cdot arctg13\pm \frac{1}{2}\cdot arccos\frac{7}{\sqrt{170}}+\pi n,\; n\in Z}$