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.
how do i solve this equation algebraicly
solve the equation 1+25cos(x)=25sin^2(x)
4 answers
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)
-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
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.
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.