Question

1) cos^4x=cos2x 2) 2sin^2x+cos4x=0

Answer 1

Ответ. (sin(x))^4+(cos(x))^4=(cos(2*x))^2+1/4; (sin(x))^4+(cos(x))^4=((cos(x))^2-(sin(x))^2)^2+1/4; 0=-2*(cos(x))^2*(sin(x))^2+1/4; -0,5*(sin(2*x))^2+0,25=0; (sin(2*x))^2=0,5; sin(2*x1)=(0,5)^0,5; sin(2*x2)=-(0,5)^0,5;

Answer 2

Let's solve each equation separately:

1. cos^4x = cos2x

We can use the double-angle identity for cosine to rewrite cos2x:

cos^4x = cos^2(2x)

Now, we can use the identity cos^2θ = (1 + cos2θ)/2:

(1 + cos2x)/2 = cos^2(2x)

Expanding cos^2(2x) using the double-angle identity for cosine:

(1 + cos2x)/2 = (1 + cos4x)/2

Multiplying both sides by 2:

1 + cos2x = 1 + cos4x

Subtracting 1 from both sides:

cos2x = cos4x

Now, we have cos2x = cos4x. To find the solutions, we can consider the possible values of x that satisfy this equation.

One possible solution is x = 0, as cos(2 * 0) = cos(4 * 0) = cos(0) = 1.

To find other solutions, we can use the periodicity of the cosine function. The cosine function has a period of 2π. Therefore, if cos2x = cos4x, then 2x and 4x must differ by an integer multiple of 2π.

Setting 4x - 2x = 2πn, where n is an integer, we have:

2x = 2πn

Dividing both sides by 2:

x = πn

So, the general solution is x = πn, where n is an integer.

2. 2sin^2x + cos4x = 0

We can use the double-angle identity for cosine to rewrite cos4x:

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

Expanding (1 - 2sin^2x)^2:

2sin^2x + 1 - 4sin^2x + 4sin^4x = 0

Combining like terms:

4sin^4x - 2sin^2x + 1 = 0

This is a quadratic equation in terms of sin^2x. Let's substitute y = sin^2x:

4y^2 - 2y + 1 = 0

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

y = (-b ± √(b^2 - 4ac))/(2a)

In this case, a = 4, b = -2, and c = 1. Substituting these values into the quadratic formula:

y = (-(-2) ± √((-2)^2 - 4 * 4 * 1))/(2 * 4) = (2 ± √(4 - 16))/(8) = (2 ± √(-12))/(8)

Since the term inside the square root is negative, there are no real solutions for y. Therefore, there are no real solutions for sin^2x, and consequently, there are no real solutions for x in the equation 2sin^2x + cos4x = 0.

In summary:

1. The solutions for cos^4x = cos2x are x = πn, where n is an integer.
2. There are no real solutions for 2sin^2x + cos4x = 0.