To simplify the equation, we can use the trigonometric identities:

Using these identities, we can rewrite the equation as:

6sin^2(x) - 11cos(x) - 10 = 0

Let y = sin(x)

6y^2 - 11√(1 - y^2) - 10 = 0

Let z = 1 - y^2

6(1 - z) - 11√z - 10 = 0

6 - 6z - 11√z - 10 = 0

-6z - 11√z - 4 = 0

Let u = √z

-6u^2 -11u - 4 = 0

Solving this quadratic equation for u, we get:

u = (-(-11) ± √((-11)^2 - 4(-6)(-4))) / (2*(-6))
u = (11 ± √(121 - 96)) / -12
u = (11 ± √25) / -12

u = (11 ± 5) / -12

u1 = (11 + 5) / -12 = 16 / -12 = -4/3
u2 = (11 - 5) / -12 = 6 / -12 = -1/2

Since z = u^2:

z1 = (-4/3)^2 = 16/9
z2 = (-1/2)^2 = 1/4

Since z = 1 - y^2:

1 - y^2 = 16/9 (or) 1 - y^2 = 1/4

y^2 = 1 - 16/9 (or) y^2 = 1 - 1/4

y^2 = 9/9 - 16/9 (or) y^2 = 4/4 - 1/4

y^2 = -7/9 (or) y^2 = 3/4

Since y^2 cannot be negative, the equation y^2 = -7/9 has no solution.

y = ± √(3/4)
y = ± √(3)/2

Therefore, the solutions to the equation are:

y = √(3)/2 and y = -√(3)/2

Since y = sin(x):

sin(x) = √(3)/2, sin(x) = -√(3)/2

The solutions for x are:

x = π/3 + 2πn, x = 2π/3 + 2πn
Where n is an integer.