Hi! Can someone help me with these? I just learned about them yesterday but I'm really confused. All my teacher wants me to do is solve each equation and that's all the directions she gave. Thanks so much!
1.) 3cot^2 x-1=0
Location on Unit Circle:
Period:
General Solution:
2.) sin^2x=3cos^2x
Location on Unit Circle:
Period:
General Solution:
I would "solve" the equation this way:
3 cot^2 x -1 = 0
cot^2 x = 1/3
tan^2 x = 3
tan x = ±√3
so x is in any of the 4 quadrants,
x = 60° , 180°-60° , 180°+60° , 360°-60°
= 60° , 120° , 240° or 300°
in radians: π/3, 2π/3, 4π/3, 5π/3
location of a unit circle:
(±1 , ±√3)
the period of tanx = 180°
notice the diffence between the 1st and 3rd solution is 180°, and the difference between the 2nd and 4th is 180°
so a general solution could be written as
(60° + 180k°) , (120° + 180k°)
I will let you change those to radians if needed.
for the #2, I will start you off
sin^2 x = 3cos^2 x
sin^2 x/cos^2 z = 3
tan^2 x = 3
tanx = ± √3
mmmhhh , looks familiar
Sure! I'd be happy to help you with these equations. Let's start with the first one:
1.) 3cot^2 x-1=0
To solve this equation, we can use the identity that relates cotangent to tangent:
cot^2 x = 1/tan^2 x
So, we can rewrite the equation as:
3(1/tan^2 x) - 1 = 0
Now, let's multiply both sides of the equation by tan^2 x to get rid of the fraction:
3 - tan^2 x = 0
Rearranging the equation, we get:
tan^2 x = 3
Now, we can take the square root of both sides:
tan x = ±√3
To find the solutions, we need to determine the values of x for which the tangent is equal to ±√3. We can do this by looking at the unit circle.
The tangent is positive in the first and third quadrants, so let's focus on those two quadrants. In the first quadrant, the reference angle is π/3, and in the third quadrant, the reference angle is 4π/3.
So, the solutions for x in the first quadrant are:
x = π/3 + 2πn, where n is an integer.
And the solutions for x in the third quadrant are:
x = 4π/3 + 2πn, where n is an integer.
This is the general solution to the equation.
Location on Unit Circle: The solutions are located at the angle π/3 in the first quadrant and 4π/3 in the third quadrant on the unit circle.
Period: The period of the tangent function is π, so the solutions repeat every π radians.
General Solution: x = π/3 + 2πn and x = 4π/3 + 2πn, where n is an integer.
Now let's move on to the second equation:
2.) sin^2x = 3cos^2x
To solve this equation, we can use the Pythagorean identity for sine and cosine:
sin^2 x + cos^2 x = 1
We can rewrite the equation as:
sin^2 x - 3cos^2 x = 0
Now, let's substitute the Pythagorean identity into the equation:
(1 - cos^2 x) - 3cos^2 x = 0
Expanding the equation, we get:
1 - cos^2 x - 3cos^2 x = 0
Simplifying further, we have:
1 - 4cos^2 x = 0
Let's rearrange the equation:
4cos^2 x = 1
Now, divide every term by 4:
cos^2 x = 1/4
Take the square root of both sides:
cos x = ±1/2
To find the solutions, we need to determine the values of x for which the cosine is equal to ±1/2. Again, let's look at the unit circle for this.
The cosine is positive in the first and fourth quadrants, so let's focus on those two quadrants. In the first quadrant, the reference angle is π/3, and in the fourth quadrant, the reference angle is 5π/3.
So, the solutions for x in the first quadrant are:
x = π/3 + 2πn, where n is an integer.
And the solutions for x in the fourth quadrant are:
x = 5π/3 + 2πn, where n is an integer.
Location on Unit Circle: The solutions are located at the angle π/3 in the first quadrant and 5π/3 in the fourth quadrant on the unit circle.
Period: The period of the cosine function is 2π, so the solutions repeat every 2π radians.
General Solution: x = π/3 + 2πn and x = 5π/3 + 2πn, where n is an integer.
I hope this helps! Let me know if you have any further questions.