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Answer (3)
x = 3 π , π Solutions for 2 cos ( x ) + 2 cos ( 2 x ) = 0 on [ 0 , 2 π ) .
To solve the equation 2 cos ( x ) + 2 cos ( 2 x ) = 0 on the interval [ 0 , 2 π ) , we'll follow these steps:
Recognize trigonometric identities:
- We know that cos ( 2 x ) = 2 cos 2 ( x ) − 1 (double-angle identity for cosine).
Substitute the identity into the equation:
2 cos ( x ) + 2 ( 2 cos 2 ( x ) − 1 ) = 0
Simplify:
2 cos ( x ) + 4 cos 2 ( x ) − 2 = 0
Rearrange terms:
4 cos 2 ( x ) + 2 cos ( x ) − 2 = 0
Divide the equation by 2:
2 cos 2 ( x ) + cos ( x ) − 1 = 0
This is now a quadratic equation in terms of cos ( x ) . Let u = cos ( x ) , so the equation becomes:
2 u 2 + u − 1 = 0
Solve the quadratic equation using the quadratic formula:
u = 2 a − b ± b 2 − 4 a c
where a = 2 , b = 1 , and c = − 1 .
Substitute the values into the formula:
u = 2 ⋅ 2 − 1 ± 1 2 − 4 ⋅ 2 ⋅ ( − 1 )
u = 4 − 1 ± 1 + 8
u = 4 − 1 ± 9
u = 4 − 1 ± 3
So, we have two possible values for u :
u 1 = 2 1
u 2 = − 1
Remembering that u = cos ( x ) , we substitute back to solve for x :
For u 1 = 2 1 :
cos ( x ) = 2 1
x = arccos ( 2 1 )
For u 2 = − 1 :[/tex]
cos ( x ) = − 1
x = π
However, we need to check if these solutions fall within the given interval [0, 2\pi)\
Checking [tex] arccos ( 2 1 ) :
arccos ( 2 1 ) ≈ 6 0 ∘
This falls within [ 0 , 2 π ) .[/tex]
Checking x = \pi\
π
This also falls within [0, 2\pi)\
So, the solutions to the equation [tex] 2 cos ( x ) + 2 cos ( 2 x ) = 0 on the interval [ 0 , 2 π ) are:
x = arccos ( 2 1 )
x = π
The solutions to the equation 2 cos x + 2 cos 2 x = 0 on the interval [ 0 , 2 π ) are x = 3 π , π , 3 5 π .
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