Solve exactly over the indicated interval:
2sinx+cscx=3, 0<theta<360
2sin+1/sin=3
2sin^2 - 3sin + 1 = 0
(2sin-1)(sin-1) = 0
sinx = 1/2 or 1
now it's a cinch, right?
sort of :/ how did you get from 1/sin=3 to 3sin+1 ?
To solve the equation 2sin(x) + csc(x) = 3 over the interval 0 < θ < 360, we need to manipulate the equation to isolate one of the trigonometric functions.
First, let's rewrite the equation using the reciprocal identity for csc(x):
2sin(x) + 1/sin(x) = 3
To simplify further, let's find a common denominator by multiplying each term by sin(x):
2sin^2(x) + 1 = 3sin(x)
Now, rearrange the equation to form a quadratic equation:
2sin^2(x) - 3sin(x) + 1 = 0
To solve this quadratic equation for sin(x), we can factor it or apply the quadratic formula. Let's use the quadratic formula:
sin(x) = (-b ± √(b^2 - 4ac)) / (2a)
For our equation, a = 2, b = -3, and c = 1.
sin(x) = (-(-3) ± √((-3)^2 - 4(2)(1))) / (2(2))
sin(x) = (3 ± √(9 - 8)) / 4
sin(x) = (3 ± √(1)) / 4
Since the square root of 1 is 1, we have two possibilities:
1. sin(x) = (3 + 1) / 4 = 4/4 = 1
2. sin(x) = (3 - 1) / 4 = 2/4 = 1/2
For the first possibility, sin(x) = 1. We know that sin(x) = 1 at x = 90 degrees or π/2 radians.
For the second possibility, sin(x) = 1/2. We know that sin(x) = 1/2 at x = 30 degrees or π/6 radians.
Thus, the solutions to the equation 2sin(x) + csc(x) = 3 over the interval 0 < θ < 360 are x = 30 degrees (π/6 radians) and x = 90 degrees (π/2 radians).