3sin^2x+4cos^x-13sinx*cosx=0

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To solve the equation 3sin^2x + 4cosx - 13sinx*cosx = 0 for x, we can use trigonometric identities to simplify the equation.

We know that sin^2x + cos^2x = 1, rearranging this equation we get:

sin^2x = 1 - cos^2x

Substitute sin^2x in the equation, we get:

3(1 - cos^2x) + 4cosx - 13sinx*cosx = 0

Expanding the equation, we get:

3 - 3cos^2x + 4cosx - 13sinx*cosx = 0

Now we can simplify further by replacing sinx*cosx with 0.5sin(2x):

3 - 3cos^2x + 4cosx - 6sin(2x) = 0

Now we have a quadratic equation in terms of cosine.

Rearranging the equation, we get:

3cos^2x - 4cosx + 6sin(2x) + 3 = 0

Let's denote cosx as t for simplicity:

3t^2 - 4t + 6sin(2x) + 3 = 0

This is a quadratic equation in terms of t, we can solve it using the quadratic formula:

t = [4 ± sqrt((-4)^2 - 43(6sin(2x) + 3))]/(2*3)

t = [4 ± sqrt(16 - 72sin(2x) - 36)]/6

t = [4 ± sqrt(-68 - 72sin(2x))]/6

Now, we can substitute t back as cosx to solve for x. Note that sin(2x) = 2sinxcosx:

cosx = [4 ± √(-68 - 72(2sinxcosx))]/6

cosx = [4 ± √(-68 - 144sinx)]/6

Now, solve for x using the arcsin function:

x = arcsin([4 ± √(-68 - 144sinx)]/6)