How do you find all solutions of the equation in the interval $[0,2pi)$ given $2sin2x-sqrt2=0$ ?

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Answer 1 · 2017-11-18T16:20:22

$pi/8 and (9pi)/8$

$2sin(2x)-sqrt(2)=0$

$sin2x=sqrt(2)/2$

$2x=sin^-1(sin2x)=sin^-1(sqrt(2)/2)=pi/4$

$2x=pi/4=>x=pi/8$

$-pi/8=pi+pi/8=(9pi)/8$

For: $color(white)(888)0 < x <=pi$

$pi/8 and (9pi)/8$

How do you find all solutions of the equation in the interval [0,2pi)[0,2pi) given 2sin2x-sqrt2=02sin2x-sqrt2=0?