Question

1)2sin^2x-3sinx+1=0 2)2sinxcosx-3cosx=-1 3)2cos^2x-3sinxcosx+sin^2x=0 Помогите плз

Answer 1

1) 2sin^2x - 3sinx + 1 = 0

To solve this equation, we can use the quadratic formula. The quadratic formula states that for an equation of the form ax^2 + bx + c = 0, the solutions for x can be found using the formula:

x = (-b ± √(b^2 - 4ac)) / (2a)

In this case, we have a = 2, b = -3, and c = 1. Plugging these values into the quadratic formula, we get:

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

Simplifying further:

x = (3 ± √(9 - 8)) / 4

x = (3 ± √1) / 4

x = (3 ± 1) / 4

So the solutions for this equation are:

x = (3 + 1) / 4 = 4 / 4 = 1

x = (3 - 1) / 4 = 2 / 4 = 1/2

Therefore, the solutions to the equation 2sin^2x - 3sinx + 1 = 0 are x = 1 and x = 1/2.

2) 2sinxcosx - 3cosx = -1

To solve this equation, we can factor out cos(x) from the left side:

cos(x)(2sin(x) - 3) = -1

Now we have two possibilities:

1. If cos(x) = 0, then the left side of the equation becomes 0, and we cannot satisfy the equation. Therefore, cos(x) cannot be 0.

2. If cos(x) ≠ 0, we can divide both sides of the equation by cos(x):

(2sin(x) - 3) = -1 / cos(x)

Simplifying further:

2sin(x) - 3 = -sec(x)

We can rewrite sec(x) as 1/cos(x):

2sin(x) - 3 = -1 / cos(x)

Multiplying both sides by cos(x):

2sin(x)cos(x) - 3cos(x) = -1

Now we have a quadratic equation in terms of sin(x) and cos(x). We can use the identity sin(2x) = 2sin(x)cos(x) to rewrite the equation:

sin(2x) - 3cos(x) = -1

Unfortunately, without further information or constraints, we cannot solve this equation explicitly. However, we can use numerical methods or approximations to find approximate solutions.

3) 2cos^2x - 3sinxcosx + sin^2x = 0

To solve this equation, we can simplify it by using the identity cos^2(x) + sin^2(x) = 1:

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

Simplifying further:

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

Combining like terms:

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

Now, we can use the identity sin(2x) = 2sin(x)cos(x) to rewrite the equation:

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

2 - sin^2(x) - 3sin(x)(sin(2x)/2) = 0

2 - sin^2(x) - (3/2)sin^2(2x) = 0

Unfortunately, without further information or constraints, we cannot solve this equation explicitly. However, we can use numerical methods or approximations to find approximate solutions.

If you have any further questions, please let me know!