Solve the given equation. (Enter your answers as a comma-separated list. Let k be any integer. Round terms to two decimal places where appropriate.)
sin(x) = √3/2
x = ?rad
30, 60, 90 triangle
sin30 = 1/2
sin 60 = sqrt 3 / 2
60 degrees * pi radians / 180 degrees
To solve the equation ....
if sinx = √3/2
then x = π/3 if x is in quadrant I
and x = π - π/3 which is 2π/3 in quadrant II
x = π/3 or x = 2π/3
since you did not state a domain, I took the usual 0 <----> 2π domain.
To solve the equation sin(x) = √3/2, we need to find the values of x such that the sine of x equals √3/2.
The sine function has a period of 2π, which means that the values of x repeat every 2π radians. So, we can find the solutions by looking at the angles in the unit circle that have a sine equal to √3/2.
From the unit circle, we know that the angles where sin(x) = √3/2 are π/3 and 2π/3.
Since the values repeat every 2π radians, we can add or subtract any multiple of 2π to these angles to get additional solutions.
Thus, the solutions for the equation sin(x) = √3/2 are:
x = π/3 + 2πk, π/3 - 2πk, 2π/3 + 2πk, 2π/3 - 2πk
where k is an integer.
Rounded to two decimal places, the solutions are approximately:
x = 1.05 rad, 1.05 rad, 2.09 rad, 2.09 rad (rounded to two decimal places).
Therefore, the solutions to the equation sin(x) = √3/2 are 1.05 rad and 2.09 rad.
To solve the equation sin(x) = √3/2, we need to find the values of x that satisfy this equation.
First, we know that sin(x) = √3/2 corresponds to the special angle π/3 or 60 degrees in the unit circle. This is because sin(π/3) = √3/2.
Since the sine function is periodic with a period of 2π (or 360 degrees), we can add or subtract multiples of 2π to find additional values:
x = π/3 + 2πk or x = -π/3 + 2πk
where k is any integer.
Now, let's convert these values to radians and round them to two decimal places:
x = π/3 + 2πk ≈ 1.05 + 2πk
x = -π/3 + 2πk ≈ -0.52 + 2πk
Note: In these equations, k represents any integer, allowing us to generate an infinite set of solutions.