Помогите, очень срочно!!!!

1) $0.04= \frac{4}{100}= \frac{1}{25}= \frac{1}{5^2}= 5^{-2} \\ ( 5^{-2}) ^{sinx cosx}= 5^{ -\sqrt{3} sinx } \\ -2sinxcosx=- \sqrt{3}sinx \\ 2sinxcosx= \sqrt{3}sinx \\ 2sinxcosx - \sqrt{3}sinx=0 \\ sinx(2cosx- \sqrt3x} )=0 \\ 1) sinx=0 \\ x= \pi k, \\ \frac{5 \pi }{2} \leq \pi k \leq 4 \pi \\ \frac{5}{2} \leq k \leq 4 \\ k={3; 4} \\ k=3 \\ x=3 \pi \\ k=4 \\ x=4 \pi$

2) $2cosx- \sqrt{3}=0 \\ 2cosx= \sqrt{3} \\ cosx= \frac{ \sqrt{3} }{2} \\ x_{1}= \frac{ \pi }{6}+2 \pi k \\ \frac{5 \pi }{2} \leq \frac{ \pi }{6}+2 \pi k \leq 4 \pi \\ \frac{5 \pi }{2}- \frac{ \pi }{6} \leq 2 \pi k \leq 4 \pi - \frac{ \pi }{6} \\ \frac{14 \pi }{6} \leq 2 \pi k \leq \frac{23 \pi }{6} \\ \frac{14 \pi }{6}* \frac{1}{2 \pi } \leq k \leq \frac{23 \pi }{6}* \frac{1}{2 \pi } \\ \frac{7}{6} \leq k \leq \frac{23}{12} \\$
$1 \frac{1}{6} \leq k \leq 1 \frac{11}{12}$
нет такого к.

$x_{2}=- \frac{ \pi }{6}+2 \pi k \\ \frac{5 \pi }{2} \leq - \frac{ \pi }{6}+2 \pi k \leq 4 \pi \\ \frac{5 \pi }{2}+ \frac{ \pi }{6} \leq 2 \pi k \leq 4 \pi + \frac{ \pi }{6} \\ \frac{16 \pi }{6} \leq 2 \pi k \leq \frac{25 \pi }{6} \\ \frac{8}{6} \leq k \leq \frac{25}{12} \\ 1 \frac{1}{3} \leq k \leq 2 \frac{1}{2} \\ k=2 \\ x_{2}= - \frac{ \pi }{6}+2 \pi *2=- \frac{ \pi }{6}+4 \pi = \frac{23 \pi }{6}$

Ответ: 3π; $\frac{23 \pi }{6}$ ; 4π