To solve this equation, we can first rewrite it using trigonometric identities.
Given equation: √3sin^2x + 2sin^2x - 1 = 2cosx
Rewrite sin^2x using cos^2x: √3(1 - cos^2x) + 2(1 - cos^2x) - 1 = 2cosx
Expand and simplify: √3 - √3cos^2x + 2 - 2cos^2x - 1 = 2cos√3 - √3cos^2x + 2 - 2cos^2x - 1 - 2cosx = √3 - √3cos^2x - 2cos^2x - 2cosx + 1 = 0
Now, we have a quadratic equation in terms of cosx. Let's solve for cosx using the quadratic formula:
a = -2, b = -2, c = -√3 + 1
cosx = (-b ± √(b^2 - 4ac)) / 2cosx = (2 ± √((-2)^2 - 4(-2)(-√3 + 1))) / (2*-2cosx = (2 ± √(4 - 8 + 8√3 - 4)) / -cosx = (2 ± √(-4 + 8√3)) / -cosx = (2 ± 2√3) / -4
cosx = (1 ± √3) / -2
Therefore, the solutions for cosx are (1 + √3) / -2 and (1 - √3) / -2.
To solve this equation, we can first rewrite it using trigonometric identities.
Given equation: √3sin^2x + 2sin^2x - 1 = 2cosx
Rewrite sin^2x using cos^2x: √3(1 - cos^2x) + 2(1 - cos^2x) - 1 = 2cosx
Expand and simplify: √3 - √3cos^2x + 2 - 2cos^2x - 1 = 2cos
√3 - √3cos^2x + 2 - 2cos^2x - 1 - 2cosx =
√3 - √3cos^2x - 2cos^2x - 2cosx + 1 = 0
Now, we have a quadratic equation in terms of cosx. Let's solve for cosx using the quadratic formula:
a = -2, b = -2, c = -√3 + 1
cosx = (-b ± √(b^2 - 4ac)) / 2
cosx = (2 ± √((-2)^2 - 4(-2)(-√3 + 1))) / (2*-2
cosx = (2 ± √(4 - 8 + 8√3 - 4)) / -
cosx = (2 ± √(-4 + 8√3)) / -
cosx = (2 ± 2√3) / -4
cosx = (1 ± √3) / -2
Therefore, the solutions for cosx are (1 + √3) / -2 and (1 - √3) / -2.