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?
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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.
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.?
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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