Question

Cos 2x + 3sin x = 2​

Answer 1

Ответ:

cos2x + 3sinx - 2 = 0

cos²x - sin²x + 3sinx - 2 =0 

1-sin²x - sin²x + 3sinx - 2 = 0

-2sin²x + 3sinx - 1 = 0 |*(-1)

2sin²x - 3sinx + 1 =0 

Обозначим: sinx=t,тогда

2t²-3t+1 =0 

D= 9-8 =1

t₁= 1, t₂ = 1/2

(1) sinx= 1

x₁= π/2+2πn, n ∈ z

(2) sinx= 1/2

x₂= (-1)^k arcsin1/2 + πk

x₂= (-1)^k π/6 + πk, k∈z

б) x₁= π/2+2πn, n ∈ z

n=1, x= π/2+2π= 5π/2 ∈ [π;5π/2]

x₂= (-1)^k π/6 + πk, k∈z

n= 2, x= (-1)² π/6 +2π = π/6+2π = 13π/6 ∈ [π;5π/2]

Объяснение:

Answer 2

To solve the equation cos(2x) + 3sin(x) = 2, we can use trigonometric identities and techniques. Let's go through the solution step by step.

First, we'll use the Pythagorean identity to rewrite sin(x) in terms of cos(x):

sin^2(x) + cos^2(x) = 1 sin^2(x) = 1 - cos^2(x) sin(x) = √(1 - cos^2(x))

Now, substitute this value of sin(x) into the given equation:

cos(2x) + 3√(1 - cos^2(x)) = 2

Next, we'll use the double angle identity for cosine:

cos(2x) = 2cos^2(x) - 1

Now replace cos(2x) with its equivalent expression:

2cos^2(x) - 1 + 3√(1 - cos^2(x)) = 2

Move the constants to the right side:

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

Combine like terms on the left side:

cos^2(x) + 3√(1 - cos^2(x)) = 3

Square both sides to eliminate the square root:

cos^4(x) + 6cos^2(x)(1 - cos^2(x)) + 9(1 - cos^2(x)) = 9

Expand and simplify:

cos^4(x) + 6cos^2(x) - 6cos^4(x) + 9 - 9cos^2(x) = 9

Combine the terms with cos^2(x):

-5cos^4(x) + 3cos^2(x) = 0

Factor out a common term:

cos^2(x)(-5cos^2(x) + 3) = 0

Now we have two possible equations to solve:

1. cos^2(x) = 0
2. -5cos^2(x) + 3 = 0

For the first equation, cos^2(x) = 0, we get:

cos(x) = 0

This gives us solutions at π/2 + kπ and 3π/2 + kπ, where k is an integer.

For the second equation, -5cos^2(x) + 3 = 0, we get:

5cos^2(x) = 3 cos^2(x) = 3/5

Taking the square root of both sides:

cos(x) = ±√(3/5)

This gives us solutions at arccos(√(3/5)) + 2kπ and -arccos(√(3/5)) + 2kπ, where k is an integer.

So, the solutions for the given equation cos(2x) + 3sin(x) = 2 are:

1. x = π/2 + kπ
2. x = 3π/2 + kπ
3. x = arccos(√(3/5)) + 2kπ
4. x = -arccos(√(3/5)) + 2kπ

where k is an integer.