solve 2cos(2theta-pi)= -2 , -2 pi less or equal to theta less than 2 pi
2 cos(2Ø-π) = -2
cos (2Ø-π) = -1
2Ø-π = π
2Ø = 2π
Ø = π
the period of cos 2Ø = π , so adding/subtracting multiples of π will yield another answer
so solutions for -2π ≤ Ø ≤ 2π :
± π , ± 2π
Let x = theta
2 cos(2x - π) = -2
Divide everything by 2
cos(2x - π) = -1
Expand the left side by applying the difference of cosines:
cos(A ± B) = cos(A)cos(B) ∓ sin(A)sin(B)
Thus,
cos(2x)cos(π) + sin(2x)sin(π) = -1
cos(2x)*(-1) + sin(2x)*(0) = -1
-cos(2x) = -1
cos(2x) = 1
2x = cos^-1 (1)
x = cos^(1) / 2
Note that cos^-1 (1) = -2π or 0 or 2π
So, x = -π or 0 or π
Hope this helps~ `u`
Looks like both Jai and I both forgot an angle
I left out 0 and Jai left out 2π
Final answer:
for -2π ≤ Ø ≤ 2π
Ø = 0, ±π, ±2π
To solve the equation 2cos(2θ - π) = -2 for θ, we can follow these steps:
Step 1: Rewrite the given equation using the double-angle formula for cosine:
2cos(2θ - π) = -2
cos(2θ - π) = -1
Step 2: Use the periodicity of the cosine function to find the general solutions within the given range. The cosine function repeats every 2π radians, so we can add or subtract any multiple of 2π to find other solutions.
Step 3: Find the reference angle:
In this case, the reference angle is (2θ - π). Let's solve for it:
2θ - π = arccos(-1)
2θ - π = π
2θ = 2π
θ = π
So, the reference angle is π.
Step 4: Find the solutions within the given range:
We need to find angles θ that satisfy the equation -2π ≤ θ < 2π.
Since the reference angle is π, we can add or subtract any multiple of π to find other solutions.
The solutions within the given range are:
θ = π + 2πn, where n is an integer.
θ = -π + 2πn, where n is an integer.
So, the solutions within the given range for the equation 2cos(2θ - π) = -2 are:
θ = π + 2πn, where n is an integer, and -2π ≤ θ < 2π.