Question

5-2sin2x-6cos^2x=0

Answer 1

5-2sin2x-6cos²x=0
5sin²x+5cos²x-4sinxcosx-6cos²x=0
5sin²x-cos²x-4sinxcosx=0 (:cos²x#0)
5tg²x-4tgx-1=0
tgx=t
5t²-4t-1=0
D=16+20=36=6²
t=(4±6)/10
t1=1;t2=-1/5

1)tgx=1
x=π/4+πk;k€Z

2)tgx=-1/5
x=-arctg(1/5)+πk;k€Z

Answer 2

To solve the equation 5 - 2sin^2(x) - 6cos^2(x) = 0, we can make use of the trigonometric identity:

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

Rearranging this identity, we get:

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

Now, we can substitute sin^2(x) with 1 - cos^2(x) in the original equation:

5 - 2(1 - cos^2(x)) - 6cos^2(x) = 0.

Simplifying:

5 - 2 + 2cos^2(x) - 6cos^2(x) = 0.

Now, combine like terms:

-3 - 4cos^2(x) = 0.

Next, isolate the cosine term:

4cos^2(x) = -3.

Divide both sides by 4:

cos^2(x) = -3/4.

However, the value of cos^2(x) cannot be negative because the range of the cosine function is from -1 to 1. Hence, there are no real solutions to this equation.

Therefore, the equation 5 - 2sin^2(x) - 6cos^2(x) = 0 has no real solutions.