Question

Sinx+√3sin(7pi/2 - x) + tgx=√3

Answer 1

решение уравнения смотри ниже

Answer 2

To solve the equation sin(x) + √3sin(7π/2 - x) + tan(x) = √3, we can simplify it using trigonometric identities and then find the values of x that satisfy the equation.

Let's simplify the equation step by step:

1. Replace sin(7π/2 - x) with cos(x): sin(x) + √3cos(x) + tan(x) = √3

2. Rewrite tan(x) as sin(x)/cos(x): sin(x) + √3cos(x) + sin(x)/cos(x) = √3

3. Multiply through by cos(x) to eliminate the fraction: sin(x)cos(x) + √3cos^2(x) + sin(x) = √3cos(x)

4. Rearrange the terms: sin(x)cos(x) + sin(x) = √3cos(x) - √3cos^2(x)

5. Factor out sin(x) on the left side: sin(x)(cos(x) + 1) = √3cos(x)(1 - cos(x))

6. Divide both sides by cos(x)(1 - cos(x)): sin(x)/(1 - cos(x)) = √3cos(x)/cos(x)

7. Simplify: sin(x)/(1 - cos(x)) = √3

Now, we have an equation in terms of sin(x) and cos(x). To solve it, we can square both sides to eliminate the square root:

[sin(x)/(1 - cos(x))]^2 = (√3)^2

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

sin^2(x) = 3 - 6cos(x) + 3cos^2(x)

Using the trigonometric identity sin^2(x) + cos^2(x) = 1, we can substitute sin^2(x) with 1 - cos^2(x):

1 - cos^2(x) = 3 - 6cos(x) + 3cos^2(x)

Rearrange the terms: 4cos^2(x) - 6cos(x) + 2 = 0

Divide through by 2: 2cos^2(x) - 3cos(x) + 1 = 0

Now, we have a quadratic equation in terms of cos(x). We can solve this equation by factoring or using the quadratic formula:

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

Factor the quadratic: (2cos(x) - 1)(cos(x) - 1) = 0

Setting each factor to zero gives us two possible solutions:

2cos(x) - 1 = 0 or cos(x) - 1 = 0

Solving each equation:

1. 2cos(x) - 1 = 0 2cos(x) = 1 cos(x) = 1/2 x = π/3 + 2πn or x = 5π/3 + 2πn, where n is an integer.

2. cos(x) - 1 = 0 cos(x) = 1 x = 2πn, where n is an integer.

Therefore, the solutions for the given equation sin(x) + √3sin(7π/2 - x) + tan(x) = √3 are: x = π/3 + 2πn, x = 5π/3 + 2πn, and x =