Question

Sin2+√2sinx=2cosx+√2​

Answer 1

Ответ: x₁=π/2+2πn   x₂=3π/4+2πn   x₃=5π/4+2πn.

Объяснение:

sin(2x)+√2*sinx=2*cosx+√2

2*sinx*cosx-2*cosx+√2*sinx-√2=0

2*cosx*(sinx-1)+√2*(sinx-1)=0

(sinx-1)*(2*cosx+√2)=0

sinx-1=0    

x₁=π/2+2πn        

2*cosx+√2=0

2*cosx=-√2  |÷2

cosx=-√2/2

x₂=3π/4+2πn        x₃=5π/4+2πn.

Answer 2

To solve the trigonometric equation sin^2(x) + √2sin(x) = 2cos(x) + √2, we'll need to manipulate the terms and try to express everything in terms of either sin(x) or cos(x).

Let's begin by using some trigonometric identities:

1. sin^2(x) = 1 - cos^2(x)
2. cos^2(x) = 1 - sin^2(x)
3. sin(45°) = cos(45°) = √2 / 2

Now, let's substitute sin^2(x) in terms of cos^2(x) using identity 2: 1 - cos^2(x) + √2sin(x) = 2cos(x) + √2

Now, move all the terms to one side: 1 - cos^2(x) - 2cos(x) + √2sin(x) - √2 = 0

Rearrange the equation: -cos^2(x) - 2cos(x) + √2sin(x) + 1 - √2 = 0

Now, let's solve for cos(x) using the quadratic formula: cos(x) = [-b ± √(b^2 - 4ac)] / 2a

where a = -1, b = -2, and c = -(√2sin(x) + 1 - √2)

cos(x) = [2 ± √((-2)^2 - 4(-1)(-(√2sin(x) + 1 - √2)))] / 2(-1)

cos(x) = [2 ± √(4 - 4(√2sin(x) + 1 - √2))] / (-2)

cos(x) = [2 ± √(4 - 4√2sin(x) - 4 + 4√2)] / (-2)

cos(x) = [2 ± √(4√2 - 4√2sin(x))] / (-2)

cos(x) = [2 ± 2√(√2 - √2sin(x))] / (-2)

cos(x) = [1 ± √(2 - 2sin(x))] / (-1)

Now, we know that cos(45°) = √2 / 2, so we can set our expression for cos(x) equal to √2 / 2:

[1 ± √(2 - 2sin(x))] / (-1) = √2 / 2

Now, solve for sin(x):

1 ± √(2 - 2sin(x)) = -√2 / 2

Let's first solve for the positive case:

1 + √(2 - 2sin(x)) = -√2 / 2

√(2 - 2sin(x)) = -√2 / 2 - 1

Square both sides to eliminate the square root:

2 - 2sin(x) = (√2 / 2 + 1)^2

2 - 2sin(x) = (1 + √2)^2

2 - 2sin(x) = 1 + 2√2 + 2

Now, solve for sin(x):

2sin(x) = 2 - (1 + 2√2 + 2)

2sin(x) = -1 - 2√2

sin(x) = (-1 - 2√2) / 2

sin(x) = -1/2 - √2

Now, let's solve for the negative case:

1 - √(2 - 2sin(x)) = -√2 / 2

√(2 - 2sin(x)) = 1 - √2 / 2

Square both sides:

2 - 2sin(x) = (1 - √2 / 2)^2

2 - 2sin(x) = (2 - √2)^2 / 4

2 - 2sin(x) = (4 - 4√2 + 2) / 4

2 - 2sin(x) = (6 - 4√2) / 4

2 - 2sin(x) = 3/2 - √2

2sin(x) = 2 - 3/2 + √2

2sin(x) = 1/2 + √2

sin(x) = (1/2 + √2) / 2

sin(x) = 1/4 + √2 / 2

So, the two solutions for sin(x) are: sin(x) = -1/2 - √2 and sin(x) = 1/4 + √2 / 2

Note: Please double-check these solutions, as there might be potential errors in the calculations. Additionally, it's important to consider the domain of the trigonometric functions when solving equations like this. Some values of x might not be valid solutions depending on the given domain.