Directions: Find the Location on Unit Circle, Period & General Solution for this problem>>>>>>> sin^2x=3cos^2x

What I have so far:

sin^2x=3cos^2
sin^2x/cos^2=3
tan^2x=3
tanx=+and- sqrt3

If you haven't already done so, you must memorize the basic trig rations of the 30-6-90° right-angled triangle, as well as the 45-45-90 triangle

I often make a quick sketch of those and it is then easy to get those trig ratios
you should recognize that tan60° = √3
and since tanx = ±√3 , you also know from the CAST rule that x must be in all 4 quadrants
so x = 60, 120, 240, and 300°
or
π/3, 2π/3, 4π/3 and 5π/3

the location of those angles are
(1,√3) , (-1,√3) ..... (you find the other two)

the period of tanx is π
so general solutions:
π/3 + kπ
2π/3 + kπ , where k is an integer

To find the location on the unit circle, you need to determine the angle whose tangent is equal to the square root of 3. The square root of 3 is an irrational number approximately equal to 1.732. We know that the tangent function is positive in the first quadrant and the third quadrant.

In the first quadrant, the tangent of an angle is positive. We can use an inverse tangent function to find the angle whose tangent is equal to square root of 3. Therefore, we have:

θ = arctan(sqrt(3))
θ ≈ 60 degrees or π/3 radians

In the third quadrant, the tangent of an angle is also positive. To find the angle in the third quadrant, we can subtract π radians (180 degrees) from the angle in the first quadrant. Therefore:

θ = π - arctan(sqrt(3))
θ ≈ 120 degrees or 2π/3 radians

Now let's determine the period of the function sin^2(x) = 3cos^2(x). The period of a trigonometric function is the length of time it takes for the function to repeat itself. Since sine and cosine have a period of 2π, we can see that the period of sin^2(x) = 3cos^2(x) will be half of that, or π.

To find the general solution for the equation sin^2(x) = 3cos^2(x), we can use the information we have gathered. In the first quadrant, the solution is θ = π/3. In the third quadrant, the solution is θ = 2π/3. Since the period is π, we can add or subtract any multiple of π to these angles to find the general solution.

Therefore, the general solution is:

x = π/3 + nπ for n = 0, 1, 2, 3...
x = 2π/3 + nπ for n = 0, 1, 2, 3...

These solutions will give you the location on the unit circle, the period, and the general solution for the equation sin^2(x) = 3cos^2(x).