Question

Sin²x-4sin x×cos x+3 cos² x=0

Answer 1

Ответ:

$x_1 = \frac{\pi}{4} + \pi k;\ \ \ x_2= \arctan (\frac{1}{3} ) + \pi k, k \in \mathbb {Z}$

Пошаговое объяснение:

$\sin ^2 x - 4\sin x \cos x + 3 \cos^2 x = 0 \ | : \ \sin^2 x \neq 0; \ x \neq \pi k, k \in \mathbb {Z}\\ 1 - 4\tan x + 3\tan^2 x = 0\\3\tan^2 x - 4\tan x + 1 = 0\ | \ t = \tan x\\3t^2-4t+1=0\\t_1=1 \ ; t_2 = \frac{1}{3} \\\tan x = 1 \ \ \ \ \tan x = \frac{1}{3}\\x_1 = \frac{\pi}{4} + \pi k;\ \ \ x_2= \arctan (\frac{1}{3} ) + \pi k, k \in \mathbb {Z}$

Answer 2

To solve the equation sin²x - 4sin(x)cos(x) + 3cos²x = 0, we can use trigonometric identities to simplify the equation and then solve for x.

Step 1: Use the identity sin²x + cos²x = 1 Replace sin²x with (1 - cos²x) in the equation: (1 - cos²x) - 4sin(x)cos(x) + 3cos²x = 0

Step 2: Rearrange the equation: 1 - cos²x - 4sin(x)cos(x) + 3cos²x = 0

Step 3: Combine like terms: 4cos²x - 4sin(x)cos(x) + 1 = 0

Step 4: Factor the quadratic equation: Let's assume t = cos(x): 4t² - 4t + 1 = 0

Step 5: Solve the quadratic equation for t: Using the quadratic formula, t = [ -b ± √(b² - 4ac) ] / 2a, where a = 4, b = -4, and c = 1: t = [ -(-4) ± √((-4)² - 4 * 4 * 1) ] / (2 * 4) t = (4 ± √(16 - 16)) / 8 t = (4 ± √0) / 8 t = (4 ± 0) / 8 t = 4 / 8 t = 1/2

Step 6: Find the values of x: Since t = cos(x), we have: cos(x) = 1/2

Now, we need to find the angles whose cosine is 1/2. These angles are π/3 and 5π/3 (along with their multiples due to the periodic nature of trigonometric functions).

So, the solutions for x are: x = π/3 + 2πn (where n is an integer) x = 5π/3 + 2πn (where n is an integer)

These are the general solutions to the equation. If you need specific solutions within a certain range, you can plug in the values of n accordingly.