Find all solutions to the given equation in the interval 0 2
Solution
(a)
2sin2x - 3sinx + 1 = 0
(2sinx - 1)(sinx - 1)
Back to the original equation
(2sinx - 1)(sinx - 1) = 0
Now use the zero property and set each set of parenthesees equal to 0
2sinx - 1 =0
2 sinx = 1
sinx = 1/2
When does sin x = 1/2 (Well in the first quadrant at pi/6 or 30 degress... does it equal 1/2 anywhere else?)
sinx - 1 = 0
sinx = 1
Where does sin x = 1? (At 90 degrees or pi/2 along with 3pi/2, 5pi/2, 7pi/2, etc etc)
(b) csc2x - cscx - 2 = 0 0° x 360°
Let u = cscx
u² - u - 2 = 0
(u - 2)(u + 1) = 0
u = 2, -1
(You could use the quadratic equation instead of factoring.)
Now reverse the substitution and solve each of these for x:
cscx = 2
sinx = ½
x = { ..., 30°, 150°, ... }
cscx = -1
sinx = -1
x = { ..., 270°, ... }
Answer:
x = { 30°, 150°, 270° }
