To solve this equation, let's rewrite it in terms of sin(2x):
Let sin(2x) = y
The equation becomes:
3y^2 + 10y + 3 = 0
This is a quadratic equation in terms of y. To solve for y, we can use the quadratic formula:
y = (-b ± √(b^2 - 4ac)) / 2a
Where a = 3, b = 10, and c = 3.
Plugging in these values, we get:
y = (-10 ± √(10^2 - 4 3 3)) / 2 * y = (-10 ± √(100 - 36)) / y = (-10 ± √64) / y = (-10 ± 8) / 6
This gives us two possible solutions for y:
y = (-10 + 8) / 6 = -2 / 6 = -1/y = (-10 - 8) / 6 = -18 / 6 = -3
Since sin(2x) cannot be outside the range of [-1, 1], the solution y = -3 is not valid.
Thus, sin(2x) = -1/3.
Now, we need to find the values of x that satisfy this:
sin(arcsin(-1/3)) = -1/3
Thus, the solution is:
2x = arcsin(-1/3) + 2πx = (arcsin(-1/3) + 2πn) / 2
Where n is an integer.
To solve this equation, let's rewrite it in terms of sin(2x):
Let sin(2x) = y
The equation becomes:
3y^2 + 10y + 3 = 0
This is a quadratic equation in terms of y. To solve for y, we can use the quadratic formula:
y = (-b ± √(b^2 - 4ac)) / 2a
Where a = 3, b = 10, and c = 3.
Plugging in these values, we get:
y = (-10 ± √(10^2 - 4 3 3)) / 2 *
y = (-10 ± √(100 - 36)) /
y = (-10 ± √64) /
y = (-10 ± 8) / 6
This gives us two possible solutions for y:
y = (-10 + 8) / 6 = -2 / 6 = -1/
y = (-10 - 8) / 6 = -18 / 6 = -3
Since sin(2x) cannot be outside the range of [-1, 1], the solution y = -3 is not valid.
Thus, sin(2x) = -1/3.
Now, we need to find the values of x that satisfy this:
sin(arcsin(-1/3)) = -1/3
Thus, the solution is:
2x = arcsin(-1/3) + 2π
x = (arcsin(-1/3) + 2πn) / 2
Where n is an integer.