Hi! If someone can help me with these, that would be great! Thanks!
Directions: Solve each equation.
1.) 2cosx+sqrt2=0
Location on Unit Circle:
Period:
General Solution:
2.) 3cot^2x-1=0
Location on Unit Circle:
Period:
General Solution:
I'll do one, and you follow the steps for the other.
2cosx + √2 = 0
cosx = -√2/2 = -1/√2
x = π/4
on the unit circle, x = cosθ and y = sinθ, so the location is (-1/√2,1/√2) or (-1/√2,-1/√2) (QII or QIII)
as usual, cosx has period 2π
x = (2k+1)π ± π/4
Of course! I'll be happy to help you solve these equations.
1.) To solve the equation 2cos(x)+√2=0, we can start by isolating the cosine term. Subtracting √2 from both sides of the equation gives us:
2cos(x) = -√2
Next, we want to find the value of x that satisfies this equation. We can use the inverse cosine function to find the angle for which the cosine value is equal to -√2/2.
On the unit circle, the cosine value at the special angles 45° and 315° is -√2/2. These angles are in the first and fourth quadrants, respectively.
Since the cosine function has a period of 360°, we can find other solutions by adding multiples of 360° to our initial solutions.
Therefore, the location on the unit circle is x = 45° + 360°n and x = 315° + 360°n, where n is an integer.
The period of the cosine function is 360°, so the values of x repeat every 360°.
The general solution is x = 45° + 360°n and x = 315° + 360°n, where n is an integer.
2.) For the equation 3cot^2(x) - 1 = 0, we can start by isolating the cotangent term. Adding 1 to both sides of the equation gives us:
3cot^2(x) = 1
Next, we want to find the value of x that satisfies this equation. We can use the inverse cotangent function to find the angle for which the cotangent value is equal to 1/√3.
On the unit circle, the cotangent value is equal to 1/√3 at the special angles 30° and 150°. These angles are in the first and second quadrants, respectively.
Since the cotangent function has a period of 180°, we can find other solutions by adding multiples of 180° to our initial solutions.
Therefore, the location on the unit circle is x = 30° + 180°n and x = 150° + 180°n, where n is an integer.
The period of the cotangent function is 180°, so the values of x repeat every 180°.
The general solution is x = 30° + 180°n and x = 150° + 180°n, where n is an integer.
I hope this helps! Let me know if you have any other questions.