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Sin(x)=2
Solve for x
How do you solve for x?

User SVD
by
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1 Answer

8 votes

Answer:

There is no real solution for this equation. However, the solution exists in complex world.


We have,\\e^(ix) = cosx + isinx.......(1)\\e^(-ix) = cos(-x) + isin(-x)\\or, e^(-ix) = cosx - isin(x)......(2)\\Subtracting equation(2) from equation (1),\\e^(ix) - e^(-ix) = 2isin(x)\\or, sin(x) = (e^(ix)-e^(-ix))/(2i)\\We need to solve:\\sin(x) = 2\\or, (e^(ix)-e^(-ix))/(2i) = 2\\or, e^(ix) - e^(-ix) = 4i\\or, e^(ix) - (1)/(e^(ix)) = 4i\\or, e^(2ix) - 1 = 4ie^(ix)\\or, e^(2ix) - 4ie^(ix)-1=0\\\\or, (e^(ix))^(2) + (-4i) e^(ix) - 1 = 0\\

Using quadratic formula:

User Ket
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