How do you solve $sin^2x-1=0$ and find all exact general solutions?

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Answer 1 · 2017-02-08T09:23:59

$x = (n-1/2)pi$ or $(2 n-1)pi/2$

$sin^2 x-1=0$

$(sin x -1)(sin x +1) =0$

$sin x -1 =0, sin x =1, x= 1/2 pi, 5/2 pi,...$

$sin x + 1 =0, sin x =-1, x= 3/2 pi, 7/2 pi,...$

we can say that,

$x = 1/2 pi, 3/2 pi,5/2 pi, 7/2 pi,...$

we can use arithmetic progression to solve this sequence where,

$a=1/2 pi, d =pi$ .

$T_n = 1/2 pi+(n-1)pi =(1/2+n-1)pi$

$T_n =(n-1/2)pi$ or $(2 n-1)pi/2$

Therefore it general solution for $x$ is $(n-1/2)pi$ or $(2 n-1)pi/2$ for all $n$

How do you solve sin^2x-1=0sin^2x-1=0 and find all exact general solutions?