Question 594819
How would you find exact solution of sin^2(x) - 1 = 0 in the interval (0,2pi)?
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sin^2(x) - 1 = 0
(sinx+1)(sinx-1)
..
sinx+1=0
sinx=-1
x=3π/2
..
sinx-1=0
sinx=1
x=π/2

ans:
x=π/2 and 3π/2