Solve for x in the equation 2sin^2x-1=cos^2x+sinx fo

Answer 1

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.