Find all the solutions in the interval [0,2pi).
2sin2x - sqrt(2) = 0
sin2x = sqrt(2)/2
I'm confused by the double angle. Please help.
ahh, smart calculator you got there.
the period of sin2x is pi or 180 degrees
so we have to add pi to each of our answers until we are no longer in the given domain.
so x = pi/8, 3pi/8, 9pi/8 and 11pi/8
I should have taken care of that in my first post , sorry
so continue...
sin2x =√2/2
2x = 45º or 135º
x = 22.5º or x = 67.5º
in radians: x = pi/8 or 3pi/8
But my calculator also shows 9pi/8 and 11pi/8. Why do you have to add pi?
To solve the equation 2sin2x - sqrt(2) = 0, we need to find the values of x in the interval [0, 2pi) that satisfy the equation.
To make it easier to work with, let's rewrite the equation as sin2x = sqrt(2)/2.
To understand the double angle, let's review the double-angle identity for sine:
sin(2x) = 2sin(x)cos(x)
In our equation, sin(2x) becomes sin2x. So, we can rewrite the equation as:
2sin(x)cos(x) = sqrt(2)/2
Now, we need to find the values of x in the interval [0, 2pi) for which sin(x) has a value that results in sqrt(2)/2 when multiplied with cos(x).
We know that sin(x) = sqrt(2)/2 when x is either pi/4 or 3pi/4 because those are the values for which sin(x) evaluates to sqrt(2)/2.
Now, we have two possibilities for x:
1. When x = pi/4:
- sin(x) = sin(pi/4) = sqrt(2)/2
- cos(x) = cos(pi/4) = sqrt(2)/2
2. When x = 3pi/4:
- sin(x) = sin(3pi/4) = sqrt(2)/2
- cos(x) = cos(3pi/4) = -sqrt(2)/2
Therefore, the solutions in the interval [0, 2pi) are x = pi/4 and x = 3pi/4.