Question

2sin^x+3cosx-3=0 [4pi;5pi]

Answer 1

2sin²x+3cosx-3=0

2(1-cos²x)+3cosx-3=0

-2cos²x+3cosx-1=0

2cos²x-3cosx+1=0

cosx=t€[-1;1]

2t²-3t+1=0
D=9-8=1
t=(3±1)/4
t1=1;t2=1/2
1)cosx=1
x=2πn

2)cosx=1/2
x=±arccos1/2+2πn
x=±π/3+2πn;n€Z

Answer 2

2sin²x+3cosx-3=0

2(1-cos²x)+3cosx-3=0

2-2cos²x+3cosx-3=0

2cos²x-3cosx+1=0

cosx=у∈[-1;1], тогда 2у²-3у+1=0; у₁,₂=(3±√(9-8))/4; у₁=1;у₂=1/2

cosx=1; х=2πn;  n∈Z. Рассматриваемому промежутку принадлежат корень при n=2         x=4π; остальные  не подходят.

cosx=1/2, х= ±arccos1/2+2πn;    n∈Z; х=±π/3+2πn;  n∈Z

Здесь отрезку [4π;5π]  принадлежит только корень 4π+π/3=13π/3

Answer 3

To solve the equation 2sin^2(x) + 3cos(x) - 3 = 0 on the interval [4π, 5π], we can use trigonometric identities to simplify the equation and solve for x.

Let's start by rewriting the equation using the identity sin^2(x) = 1 - cos^2(x):

2(1 - cos^2(x)) + 3cos(x) - 3 = 0

Expanding and rearranging the equation:

2 - 2cos^2(x) + 3cos(x) - 3 = 0

-2cos^2(x) + 3cos(x) - 1 = 0

Now, let's solve this quadratic equation for cos(x). We can use the quadratic formula:

cos(x) = (-b ± √(b^2 - 4ac)) / (2a)

In this case, a = -2, b = 3, and c = -1. Substituting these values into the formula:

cos(x) = (-(3) ± √((3)^2 - 4(-2)(-1))) / (2(-2))

cos(x) = (-3 ± √(9 - 8)) / (-4)

cos(x) = (-3 ± √1) / (-4)

cos(x) = (-3 ± 1) / (-4)

Now we have two possible values for cos(x):

1. cos(x) = (-3 + 1) / (-4) = -2 / (-4) = 1/2
2. cos(x) = (-3 - 1) / (-4) = -4 / (-4) = 1

Since cos(x) is the ratio of the adjacent side to the hypotenuse in a right triangle, it cannot be greater than 1 in magnitude. Therefore, we discard the second solution (cos(x) = 1) since it is outside the valid range.

Now, we need to find the corresponding values of x that satisfy cos(x) = 1/2 on the interval [4π, 5π].

To find these values, we can use the inverse cosine function (also known as arccos or cos^(-1)). The inverse cosine function returns values in the range [0, π].

The solution for cos(x) = 1/2 is x = arccos(1/2). We can calculate this value:

x = arccos(1/2) ≈ π/3

Now, let's check if x = π/3 falls within the given interval [4π, 5π]:

4π ≤ π/3 ≤ 5π

Since π/3 is greater than 4π, it does not satisfy the given interval condition.

Therefore, there are no solutions for the equation 2sin^2(x) + 3cos(x) - 3 = 0 on the interval [4π, 5π].