How do you solve ${sin}^{4} x - 1 = 0$ $sin^4x-1=0$ ?

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Answer 1 · 2016-12-02T21:08:26

The solutions are ${\frac{3 π}{2} + 2 k π, \frac{π}{2} + 2 k π}$ ${(3pi)/2+2kpi,pi/2+2kpi}$ , $kin ZZ$

Let's factorise the expression

$a^2-b^2=(a+b)(a-b)$

$sin^4-1=0$

$(sin^2+1)(sin^2-1)=0$

$(sin^2+1)(sinx+1)(sinx-1)=0$

So,

$sin^2x+1=0$ (no solution)

$sinx+1=0$ and $sinx-1=0$

$sinx=-1$ and $sinx=1$

$x=(3pi)/2+2kpi$ and $x=pi/2+2kpi$ , $k in ZZ$

How do you solve sin^4x-1=0sin4x−1=0sin^4x-1=0?