Question

6sin2Xcos2X + 12=0

Answer 1

6sin2Xcos2X + 12=0
3sin4X+12=0
3sin4X=-12
sin4X=-4
Ответ: нет решений.

Answer 2

To solve the trigonometric equation 6sin^2(X)cos^2(X) + 12 = 0, we can start by factoring out the common factor of 6:

6(sin^2(X)cos^2(X) + 2) = 0

Now, we have:

sin^2(X)cos^2(X) + 2 = 0

Next, we can try to simplify the equation further by using a trigonometric identity. The double-angle identity for sine is:

sin(2θ) = 2sin(θ)cos(θ)

Let's apply this identity:

sin^2(X)cos^2(X) = (1/2 * sin(2X))^2 = (1/4) * sin^2(2X)

Now, our equation becomes:

(1/4) * sin^2(2X) + 2 = 0

To solve for sin^2(2X), we can subtract 2 from both sides:

(1/4) * sin^2(2X) = -2

Now, multiply both sides by 4 to eliminate the fraction:

sin^2(2X) = -8

Since the sine of any angle squared cannot be negative (sine is always between -1 and 1), there are no real solutions to this equation. Hence, the equation has no real solutions for X.