Question

A) (36^sinx)^cosx = 6^sqrt2 sinx b) корни на отрезке [2pi ; 7pi/2]

Answer 1

(36^sinx)^cosx = 6^V2sinx

(6^2sinxcosx)=6^V2sinx

2sinxcosx=V2sinx

2cosx=V2

cosx=V2/2

x=+- p/4 + 2pk

Answer 2

а) $(36^{sinx})^{cosx}=6^{\sqrt2sinx} \\\ 6^{2sinxcosx}=6^{\sqrt2sinx} \\\ 2sinxcosx=\sqrt2sinx \\\ 2sinxcosx-\sqrt2sinx=0 \\\ sinx(2cosx-\sqrt2)=0$

$sinx=0 \\\ x=\pi k, k \in Z$ или $cosx=\frac{\sqrt2}{2} \\\ x=б\frac{\pi}{4}+2\pi n, n \in Z$

б) На [2П; 7П/2]:

1) $2\pi\leq \pi k \leq \frac{7\pi}{2} \\\ 2\leq k \leq 3,5 \\\ k=2;\ 3; \ \ x_1=2\pi, \ x_2=3\pi.$

2) $2\pi\leq б\frac{\pi}{4}+2\pi n \leq \frac{7\pi}{2} \\\ 2\leq б\frac{1}{4}+2n \leq 3,5$

$2,25\leq 2n \leq 3,75 \\\ 1,125\leq n \leq 1,875$ или $1,75\leq 2n \leq 3,25 \\\ 0,875\leq n \leq 1,625 \\\ n=1 \\\ x_3=\frac{9\pi}{4}$

ответ: 2П; 3П; 9П/4.