solve for x if x is in radians.
cos2(x) + 4cos(x) = -3
Is that
cos^2 x + 4cosx = -3 ??
if so, then let cosx = y to get
y^2 + 4y + 3 = 0
(y+1)(y+3) = 0
y = -1 or y = -3
so
cosx = -1 ---> x = π
or
cosx = -3 , not possible
so x = π
no, its not squared. it is (cosx)(2)
(cosx)(2) wouldn't make much sense, then it would be
2cosx and the entire equation would be
6cosx = -3
cosx = -1/2
and x = 120° or 240°
did you mean cos(2x) ?
then you can use the identity
cos(2x) = 2cos^2x - 1
cos((x) + 4cos(x) = -3
2cos^2x - 1 + 4cosx = -3
2cos^2x + 4cosx + 2 = 0
cos^2x + 2cosx + 1 = 0
(cosx + 1)^2 = 0
cosx + 1 = 0
cosx = -1
x = 180° or x = π
(which by coincidence is the same answer to the other interpretation)
okay i think that is right. thank you
To solve for x in the equation cos2(x) + 4cos(x) = -3, we can use a substitution to simplify the equation.
Let's substitute cos(x) with a variable, let's say "u". Therefore, cos(x) = u.
Substituting this into the equation, we have cos2(x) + 4u = -3.
Now, we can rewrite cos2(x) in terms of u using the double angle formula:
cos2(x) = cos2(x) - sin2(x)
cos2(x) = (1 - sin2(x))
Substituting this into the equation, we get (1 - sin2(x)) + 4u = -3.
Since cos2(x) + sin2(x) = 1, we can rewrite sin2(x) as 1 - cos2(x).
Therefore, sin2(x) = 1 - (1 - sin2(x)) = sin2(x).
Now, the equation becomes 1 - sin2(x) + 4u = -3.
Rearranging the equation, we get -sin2(x) + 4u - 4 = -4.
Multiplying by -1 to change the sign, we have sin2(x) - 4u + 4 = 4.
Dividing the equation by 4, we get (1/4)sin2(x) - u + 1 = 1.
Now, let's combine like terms and simplify further:
(1/4)sin2(x) - u = 0.
Multiplying the equation by 4, we have sin2(x) - 4u = 0.
Now, we can substitute sin2(x) with 1 - cos2(x), since sin2(x) = 1 - cos2(x).
The equation becomes 1 - cos2(x) - 4u = 0.
Substituting u = cos(x), we get 1 - cos2(x) - 4cos(x) = 0.
Rearranging the equation, we have -cos2(x) - 4cos(x) + 1 = 0.
Now, let's solve this quadratic equation for cos(x).
We can factor the quadratic expression:
-(cos(x) + 1)(cos(x) + 1) = 0.
Setting each factor equal to zero, we get two possible cases:
1) cos(x) + 1 = 0:
cos(x) = -1.
This gives us x = π radians as one possible solution.
2) cos(x) + 1 = 0:
cos(x) = -1.
This gives us x = 2π radians as another possible solution.
Therefore, the solutions for x in radians are x = π and x = 2π.