Question

1)2cos^2t + cos t =0 2)cos(pi/2-t)-sin(pi+t)=sqrt2 3)cost=-1 4)sint=-sqrt3/2

Answer 1

1) 2cos^2 t + cos t = 0
cost (2cost + 1)= 0
cos t=0 или 2cost + 1=0
 
t=П/2+Пn, или 2cost=-1
                      cost= - 0,5
                   t= П/6 + 2Пn или t=-П/6 + 2Пn, n принадлежит Z.             

Answer 2

Equation 1: 2cos^2(t) + cos(t) = 0

To solve this equation, we can first factor out the common term cos(t):

cos(t)(2cos(t) + 1) = 0

Now, we have two possibilities:

1) cos(t) = 0 2) 2cos(t) + 1 = 0

For the first possibility, cos(t) = 0, we know that cos(t) is equal to 0 at π/2 radians and 3π/2 radians. So, we can write:

t = π/2 + kπ, where k is an integer t = 3π/2 + kπ, where k is an integer

For the second possibility, 2cos(t) + 1 = 0, we can solve for cos(t):

2cos(t) = -1 cos(t) = -1/2

cos(t) is equal to -1/2 at 2π/3 radians and 4π/3 radians. So, we can write:

t = 2π/3 + 2kπ, where k is an integer t = 4π/3 + 2kπ, where k is an integer

Therefore, the solutions to the equation 2cos^2(t) + cos(t) = 0 are:

t = π/2 + kπ, where k is an integer t = 3π/2 + kπ, where k is an integer t = 2π/3 + 2kπ, where k is an integer t = 4π/3 + 2kπ, where k is an integer

Equation 2: cos(pi/2 - t) - sin(pi + t) = sqrt(2)

To solve this equation, we can use the trigonometric identities:

cos(pi/2 - t) = sin(t) sin(pi + t) = -sin(t)

Replacing these identities in the equation, we get:

sin(t) - (-sin(t)) = sqrt(2) 2sin(t) = sqrt(2) sin(t) = sqrt(2)/2

sin(t) is equal to sqrt(2)/2 at π/4 radians and 3π/4 radians. So, we can write:

t = π/4 + 2kπ, where k is an integer t = 3π/4 + 2kπ, where k is an integer

Therefore, the solutions to the equation cos(pi/2 - t) - sin(pi + t) = sqrt(2) are:

t = π/4 + 2kπ, where k is an integer t = 3π/4 + 2kπ, where k is an integer

Equation 3: cos(t) = -1

The equation cos(t) = -1 is satisfied when t = π + 2kπ, where k is an integer.

Therefore, the solution to the equation cos(t) = -1 is:

t = π + 2kπ, where k is an integer

Equation 4: sin(t) = -sqrt(3)/2

The equation sin(t) = -sqrt(3)/2 is satisfied when t = 7π/6 + 2kπ and t = 11π/6 + 2kπ, where k is an integer.

Therefore, the solutions to the equation sin(t) = -sqrt(3)/2 are:

t = 7π/6 + 2kπ, where k is an integer t = 11π/6 + 2kπ, where k is an integer

I hope this helps! Let me know if you have any further questions.