To solve the equation 6 sin x cos x - 8 cos 2x = 0, we can use trigonometric identities to simplify it.
First, note that cos 2x = 2(cos x)^2 - 1. We can substitute this into the equation:
6 sin x cos x - 8(2(cos x)^2 - 1) = 06 sin x cos x - 16(cos x)^2 + 8 = 0
Now, we can use the double angle formula sin 2x = 2sin x cos x to simplify it further:
6(2sin x cos x) - 16(cos x)^2 + 8 = 012sin x cos x - 16(cos x)^2 + 8 = 0
Now, let's set cos x = u to make substitution:
12sin x u - 16u^2 + 8 = 012sin x u = 16u^2 - 83sin x u = 4u^2 - 2
Divide this equation by u:
3sin x = 4u - 2/u
Using the identity sin x = sqrt(1 - cos^2 x), we have:
3(sqrt(1 - u^2)) = 4u - 2/u
Squaring both sides:
9(1 - u^2) = 16u^2 - 8 + 49 - 9u^2 = 16u^2 - 425u^2 = 13u = ±√(13/25)u = ±√13/5
Therefore, the solutions for cos x are ±√13/5.
To solve the equation 6 sin x cos x - 8 cos 2x = 0, we can use trigonometric identities to simplify it.
First, note that cos 2x = 2(cos x)^2 - 1. We can substitute this into the equation:
6 sin x cos x - 8(2(cos x)^2 - 1) = 0
6 sin x cos x - 16(cos x)^2 + 8 = 0
Now, we can use the double angle formula sin 2x = 2sin x cos x to simplify it further:
6(2sin x cos x) - 16(cos x)^2 + 8 = 0
12sin x cos x - 16(cos x)^2 + 8 = 0
Now, let's set cos x = u to make substitution:
12sin x u - 16u^2 + 8 = 0
12sin x u = 16u^2 - 8
3sin x u = 4u^2 - 2
Divide this equation by u:
3sin x = 4u - 2/u
Using the identity sin x = sqrt(1 - cos^2 x), we have:
3(sqrt(1 - u^2)) = 4u - 2/u
Squaring both sides:
9(1 - u^2) = 16u^2 - 8 + 4
9 - 9u^2 = 16u^2 - 4
25u^2 = 13
u = ±√(13/25)
u = ±√13/5
Therefore, the solutions for cos x are ±√13/5.