Question

sin 2t=1 [ -2п; п] Найтиде, пожалуйста, сумму корней на промежутке(( очень надо( если можно, с объяснениями(

Answer 1

cosx=t 
t^2-2t-3=0 
D=16 
t=3 t=-1 
cosx=-1 
x=+-p+2pn 
n=0 1 2 3 4....
p-p+3p-3p+5p-5p+7p=7p

Answer 2

Task: Solve the equation sin(2t) = 1 on the interval [-2π, π] and find the sum of the roots.

To solve the equation sin(2t) = 1, we need to find the values of t that satisfy this equation on the given interval [-2π, π].

Let's break down the problem step by step:

Step 1: Find the general solutions of the equation sin(2t) = 1. Step 2: Determine the values of t that fall within the given interval [-2π, π]. Step 3: Find the sum of the roots within the interval.

Step 1: Find the general solutions of the equation sin(2t) = 1.

The equation sin(2t) = 1 can be rewritten as 2t = π/2 + 2πn, where n is an integer.

To find the general solutions, we solve for t:

2t = π/2 + 2πn

Dividing both sides by 2:

t = (π/2 + 2πn)/2

Simplifying:

t = π/4 + πn

So, the general solutions of the equation sin(2t) = 1 are t = π/4 + πn, where n is an integer.

Step 2: Determine the values of t that fall within the given interval [-2π, π].

We need to find the values of t that satisfy the equation sin(2t) = 1 within the interval [-2π, π].

Substituting the general solutions t = π/4 + πn into the equation sin(2t) = 1:

sin(2(π/4 + πn)) = 1

Simplifying:

sin(π/2 + 2πn) = 1

Since sin(x) = 1 when x = π/2 + 2πk, where k is an integer, we have:

π/2 + 2πn = π/2 + 2πk

Simplifying:

2πn = 2πk

Dividing both sides by 2π:

n = k

So, the values of t that satisfy the equation sin(2t) = 1 within the interval [-2π, π] are t = π/4 + πn, where n is an integer.

Step 3: Find the sum of the roots within the interval.

To find the sum of the roots within the interval [-2π, π], we need to determine the values of n that fall within this interval.

For t = π/4 + πn to be within the interval [-2π, π], we have:

-2π ≤ π/4 + πn ≤ π

Subtracting π/4 from all sides:

-2π - π/4 ≤ πn ≤ π - π/4

Simplifying:

-8π/4 - π/4 ≤ πn ≤ 4π/4 - π/4

-9π/4 ≤ πn ≤ 3π/4

Dividing all sides by π:

-9/4 ≤ n ≤ 3/4

Since n is an integer, the values of n that satisfy -9/4 ≤ n ≤ 3/4 are -2, -1, 0, 1, 2, 3.

Substituting these values of n into the equation t = π/4 + πn, we get the corresponding values of t:

t = π/4 + π(-2) = -7π/4 t = π/4 + π(-1) = -3π/4 t = π/4 + π(0) = π/4 t = π/4 + π(1) = 5π/4 t = π/4 + π(2) = 9π/4 t = π/4 + π(3) = 13π/4

The sum of these roots is:

-7π/4 + -3π/4 + π/4 + 5π/4 + 9π/4 + 13π/4 = 18π/4 = 9π/2

Therefore, the sum of the roots of the equation sin(2t) = 1 on the interval [-2π, π] is 9π/2.

Please let me know if you need any further clarification or assistance!