Solve for x in the equation 2sin^2x-1=cos^2x+sinx fo
To solve for x in the equation:
2sin^2x - 1 = cos^2x + sinx
We can rewrite the equation using the Pythagorean identity:
2(1 - cos^2x) - 1 = cos^2x + sinx
2 - 2cos^2x - 1 = cos^2x + sinx
1 - 2cos^2x = cos^2x + sinx
Rearranging the equation:
2cos^2x + sinx = 1
We know that sin^2x + cos^2x = 1, so we can substitute sin^2x = 1 - cos^2x into the equation:
2cos^2x + √(1 - cos^2x) = 1
Solving for cos^2x:
2cos^2x + √(1 - cos^2x) = 1
2cos^2x = 1 - √(1 - cos^2x)
2cos^2x = 1 - √(sin^2x)
2cos^2x = 1 - sinx
Now, we can use the Pythagorean identity again to rewrite the cosine term in terms of sinx:
2(1 - sin^2x) = 1 - sinx
2 - 2sin^2x = 1 - sinx
2sin^2x - sinx - 1 = 0
This is a quadratic equation in terms of sinx, and we can solve for sinx using the quadratic formula:
sinx = [1 ± √(1 + 8)] / 4
sinx = [1 ± √9] / 4
sinx = [1 ± 3] / 4
sinx = 1 or sinx = -1/2
Since sinx cannot equal 1, we have sinx = -1/2.
Now, we can use sinx = -1/2 to solve for cosx in the original equation 2sin^2x - 1 = cos^2x + sinx:
2(1/4) - 1 = cos^2x - 1/2
1/2 - 1 = cos^2x - 1/2
-1/2 = cos^2x - 1/2
cos^2x = 0
cosx = ±√0
cosx = 0
Therefore, the solutions for x are x = 7π/6 and x = 11π/6.