Question

Sin^2x-2sinx*cosx=3cos^2x

Answer 1

Sin^2x-2sinx*cosx-3cos^2x=0( : cos^2x)
ОДЗ: x э R
tg^2x-2tgx-3=0
tgx=t
t^2-2t-3=0
D=16>0( 2 рдк)
t1=3  t2= -1
tgx=3                            или      tgx= -1
х=arctg3+Пn; n э Z                  x= -П/4+Пn; n э Z 

Answer 2

To solve the equation sin^2x - 2sinx*cosx = 3cos^2x, we can use some trigonometric identities to simplify it.

Let's start by using the identity sin^2x + cos^2x = 1. By rearranging this identity, we get sin^2x = 1 - cos^2x.

Substituting this into the equation, we have:

(1 - cos^2x) - 2sinx*cosx = 3cos^2x

Expanding and rearranging the equation, we get:

1 - cos^2x - 2sinx*cosx - 3cos^2x = 0

Now, let's combine like terms:

-4cos^2x - 2sinx*cosx + 1 = 0

To simplify further, we can use the identity sin(2x) = 2sinx*cosx:

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

This is a quadratic equation in terms of cosx. Let's solve it by factoring or using the quadratic formula. To do that, we rewrite it as:

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

Combining the terms:

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

Using the identity cos^2x + sin^2x = 1:

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

Now the equation becomes:

-3cos^2x - 2sinxcosx + sin^2x + 1 = 0

The equation can be further simplified as:

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

Using the identity sin^2x - 2sinxcosx + cos^2x = 1:

1 - 4cos^2x + 1 = 0

Simplifying, we get:

-4cos^2x + 2 = 0

Dividing by -2:

2cos^2x - 1 = 0

Now, we have a quadratic equation in terms of cosx. Let's solve it by factoring or using the quadratic formula:

(2cosx - 1)(cosx + 1) = 0

Setting each factor equal to zero:

2cosx - 1 = 0 or cosx + 1 = 0

Solving for cosx in the first equation:

2cosx = 1 cosx = 1/2

Taking the inverse cosine (cos^(-1)) of both sides:

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

x = π/3 + 2πn or x = -π/3 + 2πn (where n is an integer)

Solving for cosx in the second equation:

cosx = -1

Taking the inverse cosine of both sides:

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

x = π + 2πn (where n is an integer)

Therefore, the solutions for the equation sin^2x - 2sinx*cosx = 3cos^2x are:

x = π/3 + 2πn, -π/3 + 2