Question

Solve over the interval [0, 2pi]

2cos2x+2=0

Answer 1

`2\cos(2x)+2=0\implies 2[\cos(2x)+1]=0\implies \cos(2x)+1=\cfrac{0}{2} \\\\\\ \cos(2x)+1=0\implies \cos(2x)=-1\implies \cos^(-1)[\cos(2x)]=\cos^(-1)(-1) \\\\\\ 2x=\cos^(-1)(-1)\implies 2x=\pi \implies x=\cfrac{\pi }{2}`

Answer 2

Explanation:

We can start by isolating the cosine term:

2cos(2x) + 2 = 0

2cos(2x) = -2

cos(2x) = -1

Now, we can use the inverse cosine function to solve for 2x:

2x = arccos(-1) + 2πn or 2x = -arccos(-1) + 2πn, where n is an integer

arccos(-1) is π, so we have:

2x = π + 2πn or 2x = -π + 2πn, where n is an integer

Simplifying, we get:

x = (π/2) + πn or x = (-π/2) + πn, where n is an integer

However, we need to restrict the solutions to the interval [0, 2π], so we only consider values of n that make the solutions fall within that range.

For n = 0, we have:

x = π/2 or x = 3π/2

For n = 1, we have:

x = (3π/2) or x = (5π/2)

Since the second set of solutions falls outside the interval [0, 2π], we only consider the first set of solutions:

x = π/2 or x = 3π/2

Therefore, the solutions over the interval [0, 2π] are:

x = π/2 or x = 3π/2.