# Trig

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Find all the solutions in the interval [0,2pi).

2sin2x - sqrt(2) = 0

sin2x = sqrt(2)/2

• Trig -

so continue...

sin2x =√2/2
2x = 45º or 135º
x = 22.5º or x = 67.5º

in radians: x = pi/8 or 3pi/8

• Trig -

But my calculator also shows 9pi/8 and 11pi/8. Why do you have to add pi?

• Trig -

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

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