Question

Хоча б одне з 4,5 або 6

Answer 1

Відповідь:

4)

$2sin^2x+5cosx-4=0\\2(1-cos^2x)+5cosx-4=0\\2-2cos^2x+5cosx-4=0\\2cos^2x-5cosx+2=0$

замена

$cosx=t\\\\2t^2-5t+2=0\\D=25-16=9=3^2\\t_{1}=2\\t_{2}=\frac{1}{2}$

$t_{1}$не удовлетворяет, т.к -1<cosx<1

возврат

$cosx=\frac{1}{2}\\x= +-\frac{\pi}{3}+2\pi k$?], k∈Z

Пояснення:

Answer 2

1 вариант 4,5,6

4)

$\displaystyle 2sin^2x+5cosx-4=0\\\\2(1-cos^2x)+5cosx-4=0\\\\-2cos^2x+5cosx-2=0\\\\cosx=t; |t|\leq 1\\\\-2t^2+5t-2=0\\\\D=25-16=9\\\\t_{1.2}=\frac{-5 \pm 3}{-4}\\\\t_1=\frac{1}{2}; t_2=2 \\\\cosx=\frac{1}{2}; x= \pm \frac{\pi }{3}+2\pi n; n \in Z$

5)

$\displaystyle cos10x=cos^2x-sin^2x\\\\\cos10x=cos2x\\\\ cos10x-cos2x=0\\\\-2sin(\frac{10x-2x}{2})*sin(\frac{10x+2x}{2})=0\\\\sin4x=0; 4x=\pi n; x=\frac{\pi n}{4}; n \in Z\\\\sin6x=0; 6x=\pi n; x=\frac{\pi n}{6}; n \in Z$

6)

$\displaystyle sin^2x+10cos^2x-11sinx*cosx=0|:cos^2x\\\\tg^2x-11tgx+10=0\\\\\ tgx=t\\\\t^2-11t+10=0\\\\D=121-40=81\\\\t_{1.2}=\frac{11 \pm 9}{2}\\\\t_1=10; t_2=1 \\\\tgx=10; x=arctg10+\pi n; n \in Z\\\\tgx=1; x=\frac{\pi }{4}+\pi n; n \in Z$