Find, to the nearest degree, the solution set of 3sin^2x = 5 cosx+1 over the demain {x|0 degrees < or = to x which is < 360 degrees}
To find the solution set of the given equation, we can rewrite it as a quadratic equation by using the identity sin^2(x) = 1 - cos^2(x). Let's substitute sin^2(x) = 1 - cos^2(x) into the equation:
3(1 - cos^2(x)) = 5cos(x) + 1
Expanding the equation:
3 - 3cos^2(x) = 5cos(x) + 1
Rearranging the terms:
3cos^2(x) + 5cos(x) - 2 = 0
Now, we can solve this quadratic equation for cos(x). You can use the quadratic formula to find the values of cos(x):
cos(x) = (-b ± sqrt(b^2 - 4ac)) / (2a)
In this case, a = 3, b = 5, and c = -2. Substituting these values into the formula:
cos(x) = (-5 ± sqrt(5^2 - 4 * 3 * -2)) / (2 * 3)
Simplifying the equation:
cos(x) = (-5 ± sqrt(25 + 24)) / 6
cos(x) = (-5 ± sqrt(49)) / 6
cos(x) = (-5 ± 7) / 6
We have two possible solutions for cos(x):
1) cos(x) = (-5 + 7) / 6 = 2/6 = 1/3
2) cos(x) = (-5 - 7) / 6 = -12/6 = -2
Now, to find the values of x, we need to take the inverse cosine (arccos) of these values.
1) arccos(1/3) = 70.5 degrees (approximately)
2) arccos(-2) is not within the domain of 0 degrees ≤ x < 360 degrees since the range of arccos is within 0 to 180 degrees.
Therefore, the nearest degree solution set of the given equation is x = 70 degrees.