Hi,
I really need help in understanding how to solve trig equations. How do you solve this equation?
Solve the equation on the interval 0 < or equal to x < or equal to 2pi.
sinx = (square root of 3)/2
I appreciate your help.
Thank you!
you know
(a) sin π/3 = √3/2
now, in the unit circle, sin x = y/r, where r is the radius (hypotenuse)
Since r is always positive, sin x is positive when y is positive, which is in QI and QII
Since sin (π-x) = sin x, sin x = √3/2 when
x = π/3 or (π - π/3) = 2π/3
To solve the equation sin(x) = √3/2 on the interval 0 ≤ x ≤ 2π, you can use two methods: the unit circle and the inverse trigonometric functions.
Method 1: Using the Unit Circle
1. Draw the unit circle, which is a circle with a radius of 1 centered at the origin.
2. Identify the point on the unit circle where the y-coordinate is √3/2. This point is located at π/3 or 60 degrees.
3. Since the given equation is sin(x) = √3/2, you need to identify all values of x that correspond to the y-coordinate of √3/2.
4. The solutions on the interval 0 ≤ x ≤ 2π are π/3 and 5π/3.
Method 2: Using Inverse Trigonometric Functions
1. Take the inverse sine (or arcsine) of both sides of the equation: sin^(-1)(sin(x)) = sin^(-1)(√3/2).
2. The inverse sine function "undoes" the sine function, so sin^(-1)(sin(x)) = x.
3. The arcsine of √3/2 is π/3. So, x = π/3.
4. However, remember that the sine function has a period of 2π, which means there are infinitely many solutions.
5. To find all solutions on the interval 0 ≤ x ≤ 2π, you can add any multiple of 2π to the initial solution. So, the solutions are π/3 + 2πn, where n is an integer.
6. Therefore, the solutions on the interval 0 ≤ x ≤ 2π are π/3 and 5π/3.
Both methods give the same solutions: π/3 and 5π/3. These values represent the angles at which sin(x) equals √3/2 on the given interval.