Find the general solution algebraically.
2 cos^2x=7cosx-3
could it be
2 cos^2-7cos +3=0
(2cosx -1)(cosx-3)=0
To find the general solution algebraically, let's rewrite the equation in a quadratic form:
2 cos^2(x) - 7 cos(x) + 3 = 0
Now, let's factorize the quadratic equation:
(2 cos(x) - 1)(cos(x) - 3) = 0
Setting each factor equal to zero gives:
2 cos(x) - 1 = 0 or cos(x) - 3 = 0
Solving the first equation, we have:
2 cos(x) = 1
cos(x) = 1/2
Taking the inverse cosine of both sides, we get:
x = ±π/3 + 2πn (where n is an integer)
Now, let's solve the second equation:
cos(x) - 3 = 0
cos(x) = 3
However, cos(x) is never greater than 1, so this equation has no real solutions.
Therefore, the general solution to the given equation algebraically is:
x = ±π/3 + 2πn (where n is an integer)
To find the general solution algebraically for the equation 2cos^2x = 7cosx - 3, follow these steps:
Step 1: Write the equation in quadratic form. In this case, let's replace cos^2x with (1 - sin^2x) using the identity cos^2x + sin^2x = 1:
2(1 - sin^2x) = 7cosx - 3
Step 2: Expand and rearrange the equation:
2 - 2sin^2x = 7cosx - 3
2sin^2x + 7cosx - 5 = 0
Step 3: Express sin^2x in terms of cosx using the identity sin^2x = 1 - cos^2x:
2(1 - cos^2x) + 7cosx - 5 = 0
2 - 2cos^2x + 7cosx - 5 = 0
-2cos^2x + 7cosx - 3 = 0
Step 4: Solve the quadratic equation. You can either factor the quadratic expression or use the quadratic formula. Factoring may be difficult in this case, so let's use the quadratic formula:
cosx = (-b ± √(b^2 - 4ac)) / (2a)
For the equation -2cos^2x + 7cosx - 3 = 0, a = -2, b = 7, and c = -3. Substituting these values into the quadratic formula:
cosx = (-7 ± √(7^2 - 4(-2)(-3))) / (2(-2))
cosx = (-7 ± √(49 - 24)) / (-4)
cosx = (-7 ± √25) / (-4)
cosx = (-7 ± 5) / (-4)
Step 5: Simplify the solutions:
(a) cosx = (-7 + 5) / (-4)
cosx = -2 / (-4)
cosx = 1/2
(b) cosx = (-7 - 5) / (-4)
cosx = -12 / (-4)
cosx = 3
Step 6: Find the values of x. To find the values of x, we use the inverse cosine function (cos^-1):
(a) x = cos^-1(1/2) + 2πn, where n is an integer
(b) x = cos^-1(3) + 2πn, where n is an integer
Now, you have the general solution algebraically for the equation 2cos^2x = 7cosx - 3.