Question

2sin^2x-2sinxcosx=3cos^2x​

Answer 1

Дано уравнение 2sin²x-2sinxcosx=3cos²x.

Разделим обе части его на cos²x:

2tg² x - 2tg x = 3.

Делаем замену tg x = t.

Получаем квадратное уравнение 2t² - 2t - 3 = 0.

D = 4 + 24 = 28.  √D = √28 = 2√7.

t1 = (2 + 2√7)/4 = (1 + √7)/2 ≈ 1,823.

t2 = (2 - 2√7)/4 = (1 - √7)/2 ≈ -0,823.

Обратная замена:

x1 = arc tg(1 + √7)/2 + πk ≈ 1,069 + πk,

x2 = arc tg(1 - √7)/2 + πk ≈ -0,924  + πk.

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Answer 2

To solve the equation 2sin^2x - 2sinxcosx = 3cos^2x, we can use trigonometric identities to simplify and manipulate the equation.

Let's start by rewriting the equation using the identity sin^2x + cos^2x = 1:

2sin^2x - 2sinxcosx = 3(1 - sin^2x)

Expanding the right side:

2sin^2x - 2sinxcosx = 3 - 3sin^2x

Now, let's simplify further by moving all terms to one side of the equation:

2sin^2x - 2sinxcosx - 3 + 3sin^2x = 0

Combining like terms:

(2sin^2x + 3sin^2x) - 2sinxcosx - 3 = 0

5sin^2x - 2sinxcosx - 3 = 0

Now, let's factor the quadratic expression:

(5sinx - 3)(sinx + 1) = 0

This equation will be satisfied when either of the factors equals zero:

5sinx - 3 = 0 or sinx + 1 = 0

For the first equation, solving for sinx:

5sinx = 3

sinx = 3/5

For the second equation:

sinx = -1

Since sinx cannot be greater than 1 or less than -1, we disregard the solution sinx = -1.

Therefore, the solution to the original equation is sinx = 3/5.