# Thread: solving trig functions sin2x

1. ## solving trig functions sin2x

This is the question

sin2x-0.8=0 in the interval 2x [0,2pi]

so far i got this
sin2x=0.8
2x=sin inverse 0.8
2x=0.927
x=0.46

the book has two answer 0.46, and 1.11. How did the book got 1.11?

2. Originally Posted by jepal
This is the question

sin2x-0.8=0 in the interval 2x [0,2pi]

so far i got this
sin2x=0.8
2x=sin inverse 0.8
2x=0.927
x=0.46

the book has two answer 0.46, and 1.11. How did the book got 1.11?
This is the solution that lies in the second quadrant. You get it via 2x = pi - 0.927.

3. Originally Posted by mr fantastic
This is the solution that lies in the second quadrant. You get it via 2x = pi - 0.927.
will this be the case too if it was cos2x.?

4. Originally Posted by jepal
will this be the case too if it was cos2x.?
If it was cos(2x) = 0.8 then you would need to remember that 2x is in the first quadrant or the fourth quadrant ....

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# solve sin2x=0.8

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