Question

Срочно нужна помощь!

Answer 1

$\displaystyle 16^{cos^2x}+16^{sin^2x}=10\\\\16^{cos^2x}+16^{1-cos^2x}=10\\\\16^{cos^2x}+ \frac{16}{16^{cos^2x}}=10\\\\16^{cos^2x}=t\\\\t+ \frac{16}{t}=10\\\\ \frac{t^2+16-10t}{t}=0\\\\t^2-10t+16=0\\\\D=100-64=36\\\\t_{1.2}= \frac{10\pm 6}{2}\\\\t_1=8; t_2=2$

$\displaystyle 16^{cos^2x}=8\\\\(2^4)^{cos^2x}=2^3\\\\4cos^2x=3\\\cos^2x= \frac{3}{4}\\\\cosx=\pm \frac{ \sqrt{3}}{2}\\\\x_{1.2}=\pm \frac{ \pi }{6}+2 \pi n; n\in Z\\\\x_{3.4}=\pm \frac{5 \pi }{6}+2 \pi n; n\in Z$


$\displaystyle 16^{cos^2x}=2\\\\2^{4cos^2x}=2\\\\4cos^2x=1\\\\cos^2x= \frac{1}{4}\\\\cosx= \pm \frac{1}{2}\\\\x_{5.6}=\pm \frac{ \pi }{3}+2 \pi n; n\in Z\\\\x_{7.8}=\pm \frac{2 \pi }{3}+2 \pi n; n\in Z$