Question

Sin^2x-5sinx*cosx+4cos^2x=0

Answer 1

Sin^2x + 5 sinx cosx + 4cos^2x=0  | : Cos²x

tg²x + 5tgx + 4 = 0

Замена tgx = t

t² + 5t +4 = 0

По дискриминанту

D=25-4*1*4=25-16=9=3^2

x1=-5+3/2=4                     x2=-5-3/2=-4

а) tgx = -1                         б)tgx = -4

x = -π/4 + πk , k ∈Z               x = -arctg 4 + πn , n ∈Z

Answer 2

To solve the equation sin^2(x) - 5sin(x)cos(x) + 4cos^2(x) = 0, we can use some trigonometric identities to simplify it.

Let's start by using the Pythagorean identity sin^2(x) + cos^2(x) = 1. Rearranging this equation, we get:

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

Substituting this expression into the original equation, we have:

sin^2(x) - 5sin(x)cos(x) + 4(1 - sin^2(x)) = 0

Expanding and rearranging the equation, we get:

sin^2(x) - 5sin(x)cos(x) + 4 - 4sin^2(x) = 0

Combining like terms, we have:

-3sin^2(x) - 5sin(x)cos(x) + 4 = 0

Now, let's use the double-angle identity for sine:

sin(2x) = 2sin(x)cos(x)

We can rewrite the equation as:

-3sin^2(x) - 5sin(x)cos(x) + 4 = -3sin^2(x) - 5(1/2)sin(2x) + 4 = 0

Now, let's substitute sin(2x) with 2sin(x)cos(x):

-3sin^2(x) - (5/2)sin(2x) + 4 = -3sin^2(x) - (5/2)(2sin(x)cos(x)) + 4 = 0

Simplifying further, we have:

-3sin^2(x) - 5sin(x)cos(x) + 4 = -3sin^2(x) - 5sin(x)cos(x) + 4 = 0

Now, we have a quadratic equation in terms of sin(x). Let's solve it using factoring or the quadratic formula.

To simplify the equation, let's divide it by -1:

3sin^2(x) + 5sin(x)cos(x) - 4 = 0

Now, we can factor the quadratic equation:

(3sin(x) - 4)(sin(x) + 1) = 0

Setting each factor equal to zero, we have:

3sin(x) - 4 = 0 or sin(x) + 1 = 0

Solving the first equation, we get:

3sin(x) = 4

sin(x) = 4/3

To solve for x, take the inverse sine (arcsin) of both sides:

x = arcsin(4/3)

However, the value arcsin(4/3) is not defined in the range of -1 to 1, so there are no real solutions for x in the first equation.

Now, let's solve the second equation:

sin(x) + 1 = 0

sin(x) = -1

The solutions for x in this case are:

x = arcsin(-1) = -π/2 + 2πn or x = π/2 + 2πn

where n is an integer.

Therefore, the solutions to the equation sin^2(x) - 5sin(x)cos(x) + 4cos^2(x) = 0 are:

x = -π/2 + 2πn or x = π/2 + 2πn

where n is an integer.