Question #10e97

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Answer 1 · 2017-05-13T14:07:37

The solution is $S = {\frac{3}{4} π + 2 k π, \frac{7}{4} π + 2 k π}$ $S={3/4pi+2kpi,7/4pi+2kpi}$ , $k in ZZ$

We need

$sin^2x+cos^2x=1$

$sin(2x)=2sinxcosx$

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

Therefore,

$(sinx-cosx)=sqrt2$

Squaring both sides

$(sinx-cosx)^2=(sqrt2)^2$

$sin^2x-2sinxcosx+cos^2x=2$

$1-sin(2x)=2$

$sin2x=-1$

$2x=3/2pi$ or $2x=7/2pi$

$x=3/4pi+2kpi$ , $k in ZZ$

$x=7/4pi+2kpi$ , $k inZZ$