4sin^x-5sin x cos x-6cos^x=0

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To solve this trigonometric equation, we can use the double angle formula to express sin(2x) and cos(2x) in terms of sin(x) and cos(x).

Let's start by rewriting the given equation in terms of sin(2x) and cos(2x):

4sin^x - 5sin(x)cos(x) - 6cos^x = 0

Now, using the double angle formula:

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

We can express the given equation in terms of sin(2x) and cos(2x):

4sin(x)cos(x) - 5(2sin(x)cos(x)) - 6(cos^2(x) - sin^2(x)) = 0

Simplify the equation:

4sin(x)cos(x) - 10sin(x)cos(x) - 6cos^2(x) + 6sin^2(x) = 0

Combine like terms:

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

Now we can express cos^2(x) and sin^2(x) as cos(2x) and sin(2x), respectively:

cos^2(x) = (1 + cos(2x))/2
sin^2(x) = (1 - cos(2x))/2

Plug these equivalences back into the equation:

-3(1 + cos(2x)) + 3(1 - cos(2x)) - 6sin(x)cos(x) = 0

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

Divide by -6:

cos(2x) + sin(x)cos(x) = 0

Now, we can use the Pythagorean identity sin^2(x) + cos^2(x) = 1 to express sin(x):

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

Now, we have a quadratic equation in terms of cos(x). Let cos(x) = y:

1 - y^2 + y*(1-y^2) = 0

1 - y^2 + y - y^3 = 0

Rearranging terms:

-y^3 - y^2 + y + 1 = 0

Now, we can solve this cubic equation for y (cos(x)) using numerical methods or cubic formula. The solutions for cos(x) can then be used to find the corresponding values of sin(x) using sin(x) = sqrt(1 - cos^2(x)).

Hope this helps. Let me know if you need further assistance.