Question

Cos^2x+2sin^2x=5sinx*cosx

Answer 1

Решение смотрите на фотографии

Answer 2

Let's simplify the given trigonometric equation step by step:

1. Recall the Pythagorean identity: sin^2(x) + cos^2(x) = 1.

2. Divide the entire equation by cos^2(x) to eliminate the squared terms: (cos^2(x) + 2sin^2(x)) / cos^2(x) = 5sin(x) / cos(x)

3. Replace sin^2(x) in the numerator using the Pythagorean identity: (cos^2(x) + 2(1 - cos^2(x))) / cos^2(x) = 5sin(x) / cos(x)

4. Distribute and simplify in the numerator: (cos^2(x) + 2 - 2cos^2(x)) / cos^2(x) = 5sin(x) / cos(x)

5. Combine like terms in the numerator: (2 - cos^2(x)) / cos^2(x) = 5sin(x) / cos(x)

6. Now, let's express sin(x) in terms of cos(x) using the Pythagorean identity: sin^2(x) + cos^2(x) = 1 sin^2(x) = 1 - cos^2(x) sin(x) = √(1 - cos^2(x))

7. Substitute the value of sin(x) into the equation: (2 - cos^2(x)) / cos^2(x) = 5√(1 - cos^2(x)) / cos(x)

8. To simplify further, let's multiply both sides by cos^2(x): 2 - cos^2(x) = 5√(1 - cos^2(x)) * cos(x)

9. Square both sides to eliminate the square root: (2 - cos^2(x))^2 = (5√(1 - cos^2(x)) * cos(x))^2

10. Expand both sides of the equation: 4 - 4cos^2(x) + cos^4(x) = 25(1 - cos^2(x)) * cos^2(x)

11. Now, distribute and rearrange terms: cos^4(x) - 4cos^2(x) + 4 = 25cos^2(x) - 25cos^4(x)

12. Move all terms to one side of the equation: cos^4(x) + 25cos^4(x) - 4cos^2(x) - 25cos^2(x) + 4 = 0

13. Combine like terms: 26cos^4(x) - 29cos^2(x) + 4 = 0

This is a quadratic equation in terms of cos(x). Let's solve for cos(x):

Let cos^2(x) = t (to simplify notation):

26t^2 - 29t + 4 = 0

Now, factor the quadratic equation:

(13t - 2)(2t - 2) = 0

Setting each factor to zero and solving for t:

1. 13t - 2 = 0 13t = 2 t = 2/13

2. 2t - 2 = 0 2t = 2 t = 1

Recall that t = cos^2(x):

1. cos^2(x) = 2/13 cos(x) = ±√(2/13)

2. cos^2(x) = 1 cos(x) = ±1

So, the solutions for cos(x) are: cos(x) = ±√(2/13) and cos(x) = ±1.

Next, let's find the corresponding values of sin(x):

1. When cos(x) = √(2/13), use the Pythagorean identity to find sin(x): sin(x) = √(1 - cos^2(x)) sin(x) = √(1 - 2/13) sin(x) = √(11/13)

2. When cos(x) = -√(2/13): sin(x) = √(1 - cos^2(x)) sin(x) = √(1 - 2/13) sin(x) = √(11/13) [same value as in case 1, since sin(x) is positive in the first and second quadrants]

3. When cos(x) = 1, sin(x) can be any value between -1 and 1.

4. When cos(x) = -1, sin(x) can be any value between -1 and 1.

Therefore, the solutions for sin(x) are: sin(x) = √(11/13) or sin(x) can be any value between -1 and 1 when cos(x) = 1 or cos(x) = -1.

In summary, the solutions to the given trigonometric equation are:

1. cos(x) = √(2/13) and sin(x) = √(11/13)

2. cos(x) = -√(2/13) and sin(x) = √(11/13)

3. cos(x) = 1 and sin(x) can be any value between -1 and 1.

4. cos(x) = -1 and sin(x) can be any value between -1 and 1.