I need to find the exact solutions on the interval [0,2pi) for:
2sin^2(x/2) - 3sin(x/2) + 1 = 0
I would start:
(2sin(x/2)-1)(sin(x/2)-1) = 0
sin(x/2)=1/2 and sin(x/2)=1
what's next?
Ok, what angle has a sin equal to say 1/2
sin (x/2)=1
arc sin (1) = x/2
PI/2=x/2
solve for x
do the same technique for the other solution.
so the solution is pi and the other solution would be
arc sin (1/2) = x/2
pi/6 = x/2
x= pi/3
is that right?
Yes, you are on the right track!
To find the solutions for the equation (2sin(x/2)-1)(sin(x/2)-1) = 0, you correctly set each factor equal to zero and solve for x/2.
For sin(x/2) = 1/2, you can use the inverse sine function (also known as arcsine):
sin(x/2) = 1/2
x/2 = arcsin(1/2)
Now, you need to find the value of x/2. The inverse sine of 1/2 is pi/6, so:
x/2 = pi/6
Finally, you solved for x by multiplying both sides by 2:
x = (pi/6) * 2
x = pi/3
So one of the solutions is x = pi/3.
For sin(x/2) = 1, you can follow a similar process:
sin(x/2) = 1
x/2 = arcsin(1)
The inverse sine of 1 is pi/2, so:
x/2 = pi/2
Again, you can solve for x by multiplying both sides by 2:
x = (pi/2) * 2
x = pi
So the other solution is x = pi.
Therefore, the exact solutions on the interval [0, 2pi) for the equation 2sin^2(x/2) - 3sin(x/2) + 1 = 0 are x = pi/3 and x = pi.