Question

4sin^2x+cos^2x=5sinxcosx

Answer 1

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Ответ: $x = \text{arctg}\dfrac{1}{4} + \pi n, \ n \in Z; \ x= \dfrac{\pi}{4} + \pi k, \ k \in Z$

Answer 2

To solve the trigonometric equation 4sin^2(x) + cos^2(x) = 5sin(x)cos(x), we'll first use trigonometric identities to simplify the equation. The following identities will be helpful:

1. sin^2(x) + cos^2(x) = 1
2. sin(2x) = 2sin(x)cos(x)

Now, let's rewrite the equation using these identities:

4sin^2(x) + cos^2(x) = 5sin(x)cos(x) 4sin^2(x) + 1 - sin^2(x) = 5sin(x)cos(x) (Using sin^2(x) + cos^2(x) = 1) 3sin^2(x) + 1 = 5sin(x)cos(x)

Next, we can replace sin^2(x) with (1 - cos^2(x)) using the first identity:

3(1 - cos^2(x)) + 1 = 5sin(x)cos(x) 3 - 3cos^2(x) + 1 = 5sin(x)cos(x) 4 - 3cos^2(x) = 5sin(x)cos(x)

Now, we can use the double-angle identity for sin(2x) to represent sin(x)cos(x) in terms of cos(2x):

sin(2x) = 2sin(x)cos(x) 2sin(x)cos(x) = sin(2x)

So, our equation becomes:

4 - 3cos^2(x) = 5sin(x)cos(x) 4 - 3cos^2(x) = 5sin(2x)

Now, we can use the fact that 4 - 3cos^2(x) = 4(1 - cos^2(x)) = 4sin^2(x) to get:

4sin^2(x) = 5sin(2x)

Now, we have a simpler equation:

4sin^2(x) - 5sin(2x) = 0

Now, let u = sin(x), so we have:

4u^2 - 5(2u) = 0 4u^2 - 10u = 0

Factor out the common term u:

u(4u - 10) = 0

Now, we have two possibilities:

1. u = 0
2. 4u - 10 = 0

For case 1, u = 0, which means sin(x) = 0. This occurs at x = 0 and x = π (and their multiples).

For case 2, 4u - 10 = 0, so 4u = 10, and u = 10/4 = 2.5. However, the sine function's range is between -1 and 1, so there are no solutions for this case.

Thus, the solutions to the original equation 4sin^2(x) + cos^2(x) = 5sin(x)cos(x) are x = 0 and x = π (and their multiples).