Question

Sin x cos x + 2 sin^2 x = cos^2 x​

Answer 1

sinx*cosx+2sin^2x=cos^2x

поделим обе части на cos^2x получим:

tgx+2tg^2x=1

2tg^2x+tgx-1=0

пусть tgx=y

2y^2+y-1=0

D=1-4*2(-1)=9

y=-1

y=1/2

Найдем х:

1)tgx=-1

x=-pi/4+pik . k=z

2)tgx=1/2

x=arctg(1/2)+pik . k=z

Answer 2

To solve the trigonometric equation sin(x)cos(x) + 2sin^2(x) = cos^2(x), we'll use trigonometric identities to simplify the equation. Let's start:

Step 1: Use the identity sin^2(x) + cos^2(x) = 1.

Step 2: Rearrange the equation and substitute sin^2(x) with (1 - cos^2(x)).

The equation becomes: sin(x)cos(x) + 2(1 - cos^2(x)) = cos^2(x).

Step 3: Distribute the 2 on the left side: sin(x)cos(x) + 2 - 2cos^2(x) = cos^2(x).

Step 4: Move all terms to one side to set the equation to zero: sin(x)cos(x) + 2 - 2cos^2(x) - cos^2(x) = 0.

Step 5: Combine the cosine terms on the left side: sin(x)cos(x) - 3cos^2(x) + 2 = 0.

Now we have a quadratic equation in terms of cos(x). To solve this quadratic equation, we can use factoring. However, this particular equation is not easily factorable. Instead, we'll use a substitution to simplify it further:

Let z = cos(x).

Now, the equation becomes: sin(x)z - 3z^2 + 2 = 0.

Step 6: Rearrange the equation to a standard quadratic form: -3z^2 + sin(x)z + 2 = 0.

Step 7: Solve the quadratic equation for z using the quadratic formula: z = [ -sin(x) ± √(sin^2(x) - 4(-3)(2)) ] / (2(-3)).

Simplifying further: z = [ -sin(x) ± √(sin^2(x) + 24) ] / (-6).

So, we have two possible values for z in terms of sin(x).

Step 8: Solve for cos(x) using the values of z: cos(x) = z = [ -sin(x) + √(sin^2(x) + 24) ] / (-6) or cos(x) = z = [ -sin(x) - √(sin^2(x) + 24) ] / (-6).

Now, you have the expressions for cos(x) in terms of sin(x). Unfortunately, the equation does not have a simple, exact solution for x, and you can't isolate x explicitly. Instead, you can use numerical methods to find approximate solutions for specific values of sin(x).