Question

Sin^2x+3sinxcosx=4cos^2x

Answer 1

Решение
Sin^2x+3sinxcosx=4cos^2x 
Sin²x + 3sinxcosx - 4cos²x = 0 делим на cos²x ≠ 0
tg²x + 3tgx - 4 = 0
tgx = t
t² + 3t - 4 = 0
t₁ = - 4
t₂ = 1
1)  tgx = - 4
x₁ = arctg(-4) + πk, k ∈ Z
x₁ = - arctg(4) + πk, k ∈ Z

2) tgx = 1
x₂ = π/4 + πn, n ∈ Z

Answer 2

To solve the trigonometric equation sin^2(x) + 3sin(x)cos(x) = 4cos^2(x), we can use trigonometric identities to simplify and eventually find the solutions.

Let's start by using the Pythagorean identity: sin^2(x) + cos^2(x) = 1. We can rearrange this to get cos^2(x) = 1 - sin^2(x).

Now, let's substitute this value of cos^2(x) into the original equation:

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

Expand the right side:

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

Bring all the terms to one side:

4sin^2(x) + 3sin(x)cos(x) - 4 = 0.

Now, we have a quadratic equation in terms of sin(x). Let's factor it if possible:

(4sin(x) - 1)(sin(x) + 4) = 0.

Now set each factor to zero and solve for sin(x):

1. 4sin(x) - 1 = 0 4sin(x) = 1 sin(x) = 1/4

2. sin(x) + 4 = 0 sin(x) = -4 (This solution is not valid since the sine function is bounded between -1 and 1)

So, the valid solution for the original equation is:

sin(x) = 1/4.

Now, to find the corresponding values of cos(x), we can use the Pythagorean identity:

cos^2(x) = 1 - sin^2(x) cos^2(x) = 1 - (1/4)^2 cos^2(x) = 1 - 1/16 cos^2(x) = 15/16.

Taking the square root of both sides, we get two possible values for cos(x):

1. cos(x) = sqrt(15)/4
2. cos(x) = -sqrt(15)/4.

So, the solutions to the original equation are:

1. sin(x) = 1/4 and cos(x) = sqrt(15)/4
2. sin(x) = 1/4 and cos(x) = -sqrt(15)/4.