Question

Sin^4x + cos^4x = 0.5

Answer 1

$Sin^{4}x+Cos^{4}x=0,5\\\\(Sin^{2}x+Cos^{2} x)^{2}-2SinxCosx=0,5\\\\1-Sin2x=0,5\\\\Sin2x=\frac{1}{2}\\\\2x=(-1)^{n}arcSin\frac{1}{2}+\pi n,nez\\\\2x=(-1)^{n}\frac{\pi }{6}+\pi n,nez\\\\x=(-1)^{n} \frac{\pi }{12}+\frac{\pi n }{2},nez$

Answer 2

To solve the trigonometric equation sin^4(x) + cos^4(x) = 0.5, we can use some trigonometric identities and algebraic manipulation. First, we'll use the Pythagorean identity:

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

Next, we can rewrite sin^4(x) and cos^4(x) as follows:

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

Now, let's substitute these expressions into the original equation:

(sin^2(x))^2 + (cos^2(x))^2 = 0.5

We notice that this expression looks similar to the Pythagorean identity, and we can simplify it further:

(sin^2(x))^2 + (cos^2(x))^2 = sin^2(x) + cos^2(x)

Using the Pythagorean identity sin^2(x) + cos^2(x) = 1, we have:

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

Now, we can solve for (sin^2(x))^2:

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

Taking the square root of both sides:

sin^2(x) = ±√(1 - (cos^2(x))^2)

Now, let's solve for sin(x):

sin(x) = ±√(1 - (cos^2(x))^2)

Since the value of sin(x) is between -1 and 1, the value inside the square root must be non-negative:

1 - (cos^2(x))^2 ≥ 0

Solving for cos^2(x):

(cos^2(x))^2 ≤ 1

Taking the square root of both sides:

|cos^2(x)| ≤ 1

Since the cosine of any angle is always between -1 and 1, the absolute value is not necessary:

cos^2(x) ≤ 1

This inequality is always true for any angle, which means that the original equation has an infinite number of solutions. In other words, there are infinitely many values of x that satisfy the equation sin^4(x) + cos^4(x) = 0.5.