Question

8sin^2(2x)+4sin^2(4x)=5

Answer 1

x ∈ {пи*k/2-пи/12, пи*k/2+пи/12}, k ∈ Z

Answer 2

To solve the equation 8sin^2(2x) + 4sin^2(4x) = 5, we can start by simplifying the trigonometric expressions using double-angle identities.

Recall the double-angle identities for sine: sin(2θ) = 2sin(θ)cos(θ) sin^2(2θ) = (1 - cos(4θ))/2

Let's rewrite the equation using these identities: 8(1 - cos^2(2x)) + 4(1 - cos^2(4x)) = 5

Now, let's simplify further: 8 - 8cos^2(2x) + 4 - 4cos^2(4x) = 5 12 - 8cos^2(2x) - 4cos^2(4x) = 5 -8cos^2(2x) - 4cos^2(4x) = -7

Next, we can apply the double-angle identities for cosine: cos(2θ) = 2cos^2(θ) - 1 cos(4θ) = 8cos^4(θ) - 8cos^2(θ) + 1

Using these identities, we have: -8(2cos^2(x) - 1)^2 - 4(8cos^4(x) - 8cos^2(x) + 1) = -7

Expanding and simplifying: -16cos^4(2x) + 32cos^2(2x) - 8 - 32cos^4(4x) + 32cos^2(4x) - 4 = -7 -16cos^4(2x) + 32cos^2(2x) - 32cos^4(4x) + 32cos^2(4x) - 12 = 0

Let's denote cos^2(2x) as a and cos^2(4x) as b: -16a^2 + 32a - 32b^2 + 32b - 12 = 0

Now, this equation can be solved as a quadratic equation in terms of a and b. However, solving it explicitly might be quite complex, and the resulting solution might not be easily interpretable. If you have any specific values or constraints for x, I can provide a more specific solution.