Question

5 sin квадрат x -cos квадрат x = 4 + 4 sincosx

Answer 1

Решите уравнение:
1472:(7+5х)=46 

(х+413):26=103
4,7*(-4y-3)-3,5(6-2y)=0,3

3*(0,2x-5)-4*(0,3x-5)=10+0,4x

12-(x+1 3//8)=10 7//24
(х-1 5//7)+22 4//7+22 4//7=26

Answer 2

To solve the equation 5sin²x - cos²x = 4 + 4sinxcosx, we can simplify the expression on both sides of the equation and then solve for x.

Let's start by using trigonometric identities to simplify the equation. Recall the following identities:

sin²x + cos²x = 1 2sinxcosx = sin(2x)

Now, let's apply these identities to the equation:

5sin²x - cos²x = 4 + 4sinxcosx 5sin²x - (1 - sin²x) = 4 + 4sinxcosx (using sin²x + cos²x = 1) 5sin²x - 1 + sin²x = 4 + 4sinxcosx 6sin²x - 1 = 4 + 4sinxcosx 6sin²x - 4sinxcosx - 5 = 0

Now, we have a quadratic equation in terms of sinx. Let's rewrite it as follows:

6sin²x - 4sinxcosx - 5 = 0

To solve this quadratic equation, we can use the quadratic formula:

sinx = (-b ± sqrt(b² - 4ac)) / (2a)

In our case, a = 6, b = -4cosx, and c = -5. Plugging in these values, we get:

sinx = (-(-4cosx) ± sqrt((-4cosx)² - 4(6)(-5))) / (2(6)) sinx = (4cosx ± sqrt(16cos²x + 120)) / 12 sinx = (cosx ± sqrt(cos²x + 30)) / 3

Now, we have an expression for sinx in terms of cosx. We can use trigonometric identities to express cosx in terms of sinx and simplify the equation further.

cosx = sqrt(1 - sin²x)

Substituting this into our expression for sinx, we get:

sinx = (sqrt(1 - sin²x) ± sqrt(sin²x + 30)) / 3

At this point, we have a transcendental equation that cannot be solved algebraically. We can use numerical methods or approximation techniques to find the solutions for x.