Solve 2cosxsinx + sinx = 0 using unit circle.
2cosxsinx + sinx = 0
sinx(2cosx + 1) = 0
sinx = 0 or cosx = -1/2
x = 0, π, 2π or x = 2π/3, 4π/3
Well, I'll give it a shot, but I can't guarantee that my answer will be as "accurate" as you would like.
Let's see here. We have the equation 2cos(x)sin(x) + sin(x) = 0.
To solve this using the unit circle, we have to determine the values of x that satisfy the equation.
First, let's tackle the sin(x) term. Since sin(x) = 0 when x = 0, pi, 2pi, etc., we can set sin(x) = 0 and solve for cos(x).
Now, you see, the problem is that when we set sin(x) = 0, the other term 2cos(x)sin(x) also becomes 0. That means that any value of x that satisfies sin(x) = 0 will also satisfy the equation 2cos(x)sin(x) + sin(x) = 0.
So, the solution to this equation is that x can be any multiple of pi.
But hey, don't be too sad about this "simple" solution. Just think of the countless jokes you can make about pi! It's a never-ending source of laughter. Pi is practically the clown of the number world!
To solve the equation 2cosxsinx + sinx = 0 using the unit circle, follow these steps:
Step 1: Rearrange the equation
Start by rearranging the equation as follows:
sinx(2cosx + 1) = 0
Step 2: Identify possible values for sinx
Since the unit circle represents the values of sine, we know that sinx = 0 when x = nπ, where n is an integer.
Step 3: Solve for cosx
Now let's solve for cosx. We have two cases to consider:
Case 1: sinx = 0
If sinx = 0, then the first factor becomes zero, which implies that 2cosx + 1 = 0. Solving this equation, we get:
2cosx = -1
cosx = -1/2
Using the unit circle, we find the value of x when cosx = -1/2. We can see that cosx = -1/2 at two points: x = 2π/3 + 2nπ and x = 4π/3 + 2nπ, where n is an integer.
Case 2: 2cosx + 1 = 0
If 2cosx + 1 = 0, then the second factor becomes zero. Solving this equation, we get:
2cosx = -1
cosx = -1/2
Using the unit circle, we find the value of x when cosx = -1/2. We can see that cosx = -1/2 at two points: x = 2π/3 + 2nπ and x = 4π/3 + 2nπ, where n is an integer.
Step 4: Combine the solutions
Combining the solutions from both cases, we have:
x = 2π/3 + 2nπ, x = 4π/3 + 2nπ, and x = nπ
Therefore, the solution to the equation 2cosxsinx + sinx = 0 using the unit circle is:
x = 2π/3 + 2nπ, x = 4π/3 + 2nπ, and x = nπ, where n is an integer.
To solve the equation 2cosxsinx + sinx = 0 using the unit circle, we can use the trigonometric identities and properties of the unit circle to find the values of x that satisfy the equation.
First, we can factor out sinx from the equation:
sinx(2cosx + 1) = 0
Now, we have two possibilities:
1) sinx = 0
2) 2cosx + 1 = 0
Let's solve each possibility one by one:
1) sinx = 0
If sinx = 0, we know that x can be any multiple of π since sinx = 0 at x = nπ, where n is an integer.
So, x = nπ, where n is an integer.
2) 2cosx + 1 = 0
To solve this equation, let's isolate cosx first:
2cosx = -1
Dividing both sides by 2:
cosx = -1/2
Now, we need to find the value of x such that cosx = -1/2. Looking at the unit circle, we know that cosx = -1/2 when x is equal to 2π/3 or 4π/3. This is because the x-coordinate of the unit circle at these angles is -1/2.
So, the solutions for cosx = -1/2 are x = 2π/3 + 2π*n and x = 4π/3 + 2π*n, where n is an integer.
Therefore, the solutions for the original equation 2cosxsinx + sinx = 0 using the unit circle are:
x = nπ, where n is an integer (for sinx = 0)
and
x = 2π/3 + 2π*n and x = 4π/3 + 2π*n, where n is an integer (for cosx = -1/2).