2sinx-cosx/2cosx+sinx=3 найти tgx

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вопрос

To find the value of tg(x), we need to first simplify the given equation:

2sinx - cosx / 2cosx + sinx = 3

Now, let's replace sinx with y and cosx with x for simplicity:

2y - x / 2x + y = 3

Now, let's multiply both sides of the equation by (2x + y) to clear the denominator:

(2y - x)(2x + y) = 3(2x + y)

Expanding both sides:

4xy + 2y^2 - 2x^2 - x^2 = 6x + 3y
4xy - 3x^2 + 2y^2 = 6x + 3y

Next, we know that tan(x) = sin(x) / cos(x), so we can express y and x in terms of sin(x) and cos(x) respectively:

y = sin(x)
x = cos(x)

Substitute x = cos(x) and y = sin(x) in our previous equation:

4sin(x)cos(x) - 3cos^2(x) + 2sin^2(x) = 6cos(x) + 3sin(x)

Now we will use the identity sin^2(x) + cos^2(x) = 1 to simplify this equation:

4sin(x)cos(x) - 3(1 - sin(x)^2) + 2sin(x)^2 = 6cos(x) + 3sin(x)

4sin(x)cos(x) - 3 + 3sin(x)^2 + 2sin(x)^2 = 6cos(x) + 3sin(x)

4sin(x)cos(x) - 3 + 5sin^2(x) = 6cos(x) + 3sin(x)

Divide by sin(x)cos(x) to get the tg(x) term:

4 - 3/sin(x)cos(x) + 5(sin(x)/cos(x))^2 = 6/cos(x) + 3tan(x)

Now, substitute tg(x) = sin(x)/cos(x)

4 - 3tg(x) + 5(tg(x))^2 = 6/cos(x) + 3tg(x)

This is a quadratic equation in terms of tg(x). We can solve this equation by rearranging it and then using the quadratic formula.