Solve for x in the equation 2sin²x-1=cos²x+sinx
To solve for x in the equation 2sin²x-1=cos²x+sinx, we first need to use the trigonometric identity sin²x + cos²x = 1.
So, we can rewrite the equation as:
2sin²x - 1 = (1 - sin²x) + sinx
2sin²x - 1 = 1 - sin²x + sinx
2sin²x - 1 = sinx - sin²x + 1
Now, we can combine like terms and simplify the equation:
2sin²x - sinx + sin²x = 1
3sin²x - sinx - 1 = 0
Now, we have a quadratic equation in terms of sinx. Let's make it look more like a standard quadratic equation by renaming sinx as y:
3y² - y - 1 = 0
This equation can be solved using the quadratic formula:
y = (-(-1) ± √((-1)² - 4(3)(-1))) / (2(3))
y = (1 ± √(1 + 12)) / 6
y = (1 ± √13) / 6
Now that we have the possible values for sinx, we can find the corresponding values of x. The possible values for x are:
x = sin⁻¹((1 + √13) / 6)
x = sin⁻¹((1 - √13) / 6)