how do i solve this equation algebraicly
solve the equation 1+25cos(x)=25sin^2(x)
Substitute sin²(x)=1-cos²(x).
This results in a quadratic equation in cos(x). Solve for y=cos(x) and evaluate x=arccos(y) if -1≤y≤1.
1+25cos(x)=25sin^2(x)
-25 sin^2 x + 25 cos x + 1
-25 (1 - cos^2 x) + 25 cos x + 1
-25 + 25 cos^2 x + 25 cos x + 1
25 cos^2 x + 25 cos x - 24
substitute u = cos x
25u^2 + 25u - 24
complete the square
u^2 + u - 24/25 = 0
u^2 + u = 24/25
u^2 + u + 1/4 = 121/100
(u + 1/2)^2 = 121/100
take square root of both sides
| u + 1/2 | = 11/10
u + 1/2 = -11/10 and u + 1/2 = 11/10
u = -8/5 and u = 3/5
substitute back in u = cos x
cos x = -8/5 and cos x = 3/5
x = arccos (-8/5)
x = arccos (3/5)
forgot to put
I am not a tutor
I think this is right
It's almost perfect.
In the last step,
"x = arccos (-8/5)
x = arccos (3/5) "
we have to reject the first value of arccos(-8/5) because arccos(-8/5) does not exist.
For a solution between 0 and 2π, there are two values of x that satisfy arccos(3/5), i.e.
approx. 53 degrees and 360-53 degrees.
For the general solution, add 2kπ to the solutions, where k is an integer.
To solve the equation algebraically, we can start by rearranging the equation and converting it into a more familiar format. Let's go through the steps:
1. Start with the equation: 1 + 25cos(x) = 25sin^2(x)
2. Rewrite the equation using the Pythagorean Identity (sin^2(x) + cos^2(x) = 1). Replace sin^2(x) with (1 - cos^2(x)):
1 + 25cos(x) = 25(1 - cos^2(x))
3. Distribute the 25:
1 + 25cos(x) = 25 - 25cos^2(x)
4. Move all terms to one side to set the equation to zero:
25cos^2(x) + 25cos(x) - 24 = 0
5. Now we have a quadratic equation in terms of cos(x). To solve this quadratic equation, we can either factor it or use the quadratic formula. In this case, let's use factoring:
(5cos(x) - 3)(5cos(x) + 8) = 0
6. Set each factor equal to zero:
5cos(x) - 3 = 0 OR 5cos(x) + 8 = 0
7. Solve each equation separately:
For 5cos(x) - 3 = 0:
5cos(x) = 3
cos(x) = 3/5
For 5cos(x) + 8 = 0:
5cos(x) = -8
cos(x) = -8/5
8. To find the solutions for x, we can use the inverse cosine function (also known as arccos or cos^(-1)):
x = arccos(3/5) OR x = arccos(-8/5)
Make sure to use the appropriate inverse cosine function based on the range of the angle values you are working with (degrees or radians).
Thus, the algebraic solution to the equation 1 + 25cos(x) = 25sin^2(x) is:
x = arccos(3/5) OR x = arccos(-8/5)