1-8sin*cosx+6cos^2x=0

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To solve this equation, we can first rewrite it in terms of sine and cosine using the Pythagorean identity sin^2(x) + cos^2(x) = 1:

8sin(x)cos(x) + 6cos^2(x) = 0
8sin(x)cos(x) + 6(1 - sin^2(x)) = 0
8sin(x)cos(x) + 6 - 6sin^2(x) = 0

Rearranging terms:

-6sin^2(x) + 8sin(x)cos(x) + 6 = 0

Now, we can factor out a common factor of -2:

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

Now, we can rewrite the middle term in terms of sine and cosine:

-2(3sin(x)(sin(x) - 4cos(x)) - 3) = 0

Now, we see that the equation can be simplified to:

3sin(x)(sin(x) - 4cos(x)) = 3

Dividing by 3:

sin(x)(sin(x) - 4cos(x)) = 1

Now, we can use the double angle identity for sine to further simplify:

sin(x)(sin(x) - 4√(1-sin^2(x))) = 1
sin^2(x) - 4√(1-sin^2(x))sin(x) = 1

Let y = sin(x):

y^2 - 4√(1-y^2)y - 1 = 0

Now, we have a quadratic equation in terms of y, which we can solve using the quadratic formula:

y = [4√(1-y^2) +- √(16(1-y^2) + 4)] / 2

y = [4√(1-y^2) +- √(4 + 12y^2)] / 2

We can then substitute back sin(x) for y to find the solutions for sin(x).