i need help finding the smallest positive number for x that cos(x^2+5x+12)=0?
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so you desire that x² + 5x + 12 = [2n+1]π / 2 ---->
x = - 5 / 2 ± √ ( [n +0.5]π -23 / 4 )----->
[n + 0.5 ]π - 23 / 4 > 25 / 4 --->
nπ > 49/4 - π / 2---> n > [ 49 / (4π) ] - 1 / 2 =3.399 or n = 4---->
- 5 / 2 + √( 4.5 π - 23/4 ) = 0.39606059004
x = - 5 / 2 ± √ ( [n +0.5]π -23 / 4 )----->
[n + 0.5 ]π - 23 / 4 > 25 / 4 --->
nπ > 49/4 - π / 2---> n > [ 49 / (4π) ] - 1 / 2 =3.399 or n = 4---->
- 5 / 2 + √( 4.5 π - 23/4 ) = 0.39606059004