solve:
cos^4(x)-sin^4(x)=cos(2x)
cos(4x)=cos^2(x)-sin^2(x)
pls help my teacher didn't teach us anything
seems like I just did these already
Remember your double-angle formulas
cos(2θ) = cos^2(θ) - sin^2(θ)
so the first one is just an identity
cos^4(x)-sin^4(x)
= (cos^2x + sin^2x)(cos^2x - sin^2x)
= 1 * (cos^2x) - sin^2x)
= cos(2x)
as written, the next one is not an identity
cos(4x) = 2cos^2(2x) - 1
so now you have
2cos^2(2x) - 1 = cos(2x)
2cos^2(2x) - cos(2x) - 1 = 0
(2cos(2x) + 1)(cos(2x - 1) = 0
so,
cos(2x) = -1/2
cos(2x) = 1
2x = 0+2kπ
x = kπ
or
2x = 2π/3 + 2kπ or 4π/3+2kπ
x = π/3 + kπ or 2π/3 + kπ = kπ ± π/3
To solve the equation cos^4(x) - sin^4(x) = cos(2x), we can use the identity cos(2x) = cos^2(x) - sin^2(x). We will substitute this identity into the equation and then simplify:
cos^4(x) - sin^4(x) = cos^2(x) - sin^2(x)
Now, let's substitute the value of cos(2x) in place of cos^2(x) - sin^2(x):
cos^4(x) - sin^4(x) = cos(2x)
(cos^2(x) - sin^2(x))(cos^2(x) + sin^2(x)) = cos(2x)
cos^2(x) - sin^2(x) = cos(2x)
Now, let's simplify further:
Using the Pythagorean identity cos^2(x) + sin^2(x) = 1, we can replace cos^2(x) - sin^2(x) in the equation:
1 - 2sin^2(x) = cos(2x)
2sin^2(x) + cos(2x) - 1 = 0
This is a quadratic equation in sin^2(x). Let's solve it using the quadratic formula:
sin^2(x) = (-b ± √(b^2 - 4ac))/(2a)
In this case, a = 2, b = 1, and c = -1. Substituting these values into the quadratic formula:
sin^2(x) = (-(1) ± √((1)^2 - 4(2)(-1)))/(2(2))
sin^2(x) = (-1 ± √(1 + 8))/4
sin^2(x) = (-1 ± √9)/4
sin^2(x) = (-1 ± 3)/4
sin^2(x) = 2/4 or -4/4
sin^2(x) = 1/2 or -1
Taking the square root of both sides, we have:
sin(x) = ±√(1/2) or ±√(-1)
sin(x) = ±(1/√2) or ±i
From here, we have multiple solutions for x, depending on the values of sin(x). The possible values for x are based on the unit circle and the periodic nature of sine and cosine functions.
To find the exact values of x, we need to consider the different quadrants where sine is positive or negative. We can use the unit circle or trigonometric identities to find the values of x.
Overall, the equation has multiple solutions due to the periodic nature of sine and cosine functions, and x can take various values.