Question

2cosx+2=0 and 2sinx+2=0 please. Time 8 hours

Answer 1

2cosx+2=0
2cosx=-2
cosx=-1
x=π+2πn

2sinx+2=0
2sinx=-2
sinx=-1
x=3π/4+2πn

Answer 2

To solve the equations:

1. $2\cos(x) + 2 = 0$
2. $2\sin(x) + 2 = 0$

Let's start with the first equation:

1. $2\cos(x) + 2 = 0$

Subtract 2 from both sides:

$2\cos(x) = -2$

Now, divide both sides by 2:

$\cos(x) = -1$

The solution for this equation is $x = \pi + 2\pi n$, where $n$ is an integer. This is because $\cos(\pi) = -1$ and the cosine function repeats every $2\pi$ radians.

Now, let's move on to the second equation:

2. $2\sin(x) + 2 = 0$

Subtract 2 from both sides:

$2\sin(x) = -2$

Divide both sides by 2:

$\sin(x) = -1$

The solution for this equation is $x = \frac{3\pi}{2} + 2\pi n$, where $n$ is an integer. This is because $\sin\left(\frac{3\pi}{2}\right) = -1$, and the sine function also repeats every $2\pi$ radians.

So, the solutions for the two equations are:

For the first equation: $x = \pi + 2\pi n$, where $n$ is an integer.

For the second equation: $x = \frac{3\pi}{2} + 2\pi n$, where $n$ is an integer.

These solutions cover all possible values of $x$ that satisfy the given equations.