Question #7c3e7

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Answer 1 · 2017-02-20T14:30:39

$sin15$

$=sin(60-45)$

$=sin60cos45-cos60sin45$

$=sqrt3/2*1/sqrt2-1/2*1/sqrt2$

$=(sqrt3-1)/(2sqrt2)$

$cos15$

$=cos(60-45)$

$=cos60cos45+sin60sin45$

$=1/2*1/sqrt2+sqrt3/2*1/sqrt2$

$=(sqrt3+1)/(2sqrt2)$

Now given equation

$(sqrt3-1)cosx+(sqrt3+1)sinx=2$

Dividing both sides by $2sqrt2$ we get

$(sqrt3-1)/(2sqrt2)cosx+(sqrt3+1)/(2sqrt2)sinx=1/sqrt2$

$=>sin15^@cosx+cos15^@sinx=1/sqrt2$

$=>sin(x+15^@)=sin45^@$

$=>sin(x+pi/12)=sin(pi/4)$

$=>x+pi/12=npi+(-1)^npi/4" where " n in ZZ$

$=>x=npi+(-1)^npi/4-pi/12" where " n in ZZ$