Question

Cos²x/2-sin²x/2<корень3/2

Answer 1

Ответ:

1=-4⇒tgx=-4⇒x=-arctg4+πn,n∈z

a2=-1⇒tgx=-1⇒x=-π/4+πk,k∈z

Объяснение:

2sin²x+5sinxcosx+5cos²x-sin²x-cos²x=0/cos²x

tg²x+5tgx+4=0

tgx=a

a²+5a+4=0

a1+a2=-5 U a1*a2=4

Answer 2

Ответ:

2 cos^{2}x+2sin2x=3  

2cos^{2}x+4sinxcosx=3  

2cos^{2}x+4sinxcosx-3 sin^{2} x-3cos^2x=0

-cos^2x+4sinxcosx-3sin^2x=0

cos^2x-4sinxcosx+3sin^2x=0 | делим на cos^2x не ≠0

3tg^2x-4tgx+1=0

tgx=t

3t^2-4t+1=0

Если первый корень 1, то второй  \frac{c}{a}  =  \frac{1}{3}  

tgx=1  

x= \frac{p}{4}+pn , n∈z

tgx= \frac{1}{3}  

x=arctg \frac{1}{3} +pn, n∈z  

Найдем корни, это -3p/4, arctg \frac{1}{3} -p

Пошаговое объяснение:

Вроде так)

Answer 3

The inequality you provided is:

cos²x/2 - sin²x/2 < √3/2

We can simplify the left-hand side of the inequality using the identity:

cos²x/2 - sin²x/2 = cosx

Substituting this into the inequality, we get:

cosx < √3/2

To solve for x, we take the inverse cosine of both sides:

x < acos(√3/2)

Using the unit circle, we can see that acos(√3/2) is π/6 radians or 30 degrees. Therefore, the solution to the inequality is:

x < π/6 + 2πn or x > 11π/6 + 2πn

where n is an integer.