Question

Sinхcosх−2cos^2x+3sin^2x=0

Answer 1

Ответ:

(см. объяснение)

Пошаговое объяснение:

$\sin x\cos x-2\cos^2x+3\sin^2x=0$

Правильно решать это уравнение, выполнив деление на $\cos^2x$:

$\mathrm{tg}x-2+3\mathrm{tg}^2x=0$

Однако я приведу другое решение:

$\sin x\cos x-2\cos^2x+3\sin^2x=0\\\mathrm{ctg}x-2\mathrm{ctg}^2x+3=0\\2\mathrm{ctg}^2x-\mathrm{ctg}x-3=0\\2\mathrm{ctg}^2x+2\mathrm{ctg}x-3\mathrm{ctg}x-3=0\\2\mathrm{ctg}x(\mathrm{ctg}x+1)-3(\mathrm{ctg}x+1)=0\\(\mathrm{ctg}x+1)(2\mathrm{ctg}x-3)=0\\1)\\\mathrm{ctg}x+1=0\\\mathrm{ctg}x=-1\\x=-\dfrac{\pi}{4}+n\pi,\;n\in Z\\2)\\2\mathrm{ctg}x-3=0\\2\mathrm{ctg}x=3\\\mathrm{ctg}x=\dfrac{3}{2}\\x=\mathrm{arccot}\dfrac{3}{2}+k\pi,\;k\in Z$

Уравнение решено!

Answer 2

To solve the equation sin(x)cos(x) - 2cos^2(x) + 3sin^2(x) = 0, we can use some trigonometric identities to simplify it and then find the values of x that satisfy the equation.

First, let's rewrite sin^2(x) and cos^2(x) using the Pythagorean identity:

sin^2(x) = 1 - cos^2(x)

Now, substitute this into the original equation:

sin(x)cos(x) - 2cos^2(x) + 3(1 - cos^2(x)) = 0

Next, distribute the 3 on the right side:

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

Combine like terms:

sin(x)cos(x) - 5cos^2(x) + 3 = 0

Now, this is a quadratic equation in terms of cos(x). We can solve it for cos(x) using the quadratic formula:

cos(x) = [-B ± √(B² - 4AC)] / (2A)

In this case, A = -5, B = sin(x), and C = 3. Plug these values into the quadratic formula:

cos(x) = [-sin(x) ± √(sin²(x) - 4(-5)(3))] / (2(-5))

cos(x) = [-sin(x) ± √(sin²(x) + 60)] / (-10)

Now, you can simplify this expression for cos(x). However, note that the solutions for x will depend on the values of sin(x) and the value inside the square root. The solutions may be expressed in terms of sin(x), and you might need additional information or constraints to find specific values of x.