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hi im back.. solve cosx=
i am trying to solve for A and B....this question comes from a Differential equations...but this step goes back to basics...
Asin
2 answers
asked by
mary
605 views
Solve this equation fo rx in the interval 0<=x<=360
3sinxtanx=8 I would do it this way: sinxtanx = 8/3 sinx(sinx/cosx)=8/3
0 answers
asked by
Edward
979 views
Solve:
cos(2x-180) - sin(x-90)=0 my work: cos2xcos180 + sin2xsin180= sinxcos90 - sin90cosx -cos2x - sin2x= cosx -cos^2x + sin^2x
0 answers
asked by
Jess
773 views
tanx+secx=2cosx
(sinx/cosx)+ (1/cosx)=2cosx (sinx+1)/cosx =2cosx multiplying both sides by cosx sinx + 1 =2cos^2x sinx+1 =
0 answers
asked by
shan
978 views
Simplify #3:
[cosx-sin(90-x)sinx]/[cosx-cos(180-x)tanx] = [cosx-(sin90cosx-cos90sinx)sinx]/[cosx-(cos180cosx+sinx180sinx)tanx] =
1 answer
asked by
Anonymous
1,072 views
Solve for [0, 360)
2sinxcosx + cosx =0 2sinxcosx = -cosx 2sinx = -cosx/cosx sinx = -1/2 {210, 330) Is this correct?
1 answer
asked by
Sara
1,085 views
hey, i would really appreciate some help solving for x when:
sin2x=cosx Use the identity sin 2A = 2sinAcosA so: sin 2x = cos x
0 answers
asked by
elle
715 views
solve each equation for 0=/<x=/<2pi
sin^2x + 5sinx + 6 = 0? how do i factor this and solve? 2sin^2 + sinx = 0 (2sinx - 3)(sinx
7 answers
asked by
sh
947 views
I have a question relating to limits that I solved
lim(x-->0) (1-cosx)/2x^2 I multiplied the numerator and denominator by
1 answer
asked by
Alex
691 views
Trigonometric Identities
Prove: (tanx + secx -1)/(tanx - secx + 1)= tanx + secx My work so far: (sinx/cosx + 1/cosx +
0 answers
asked by
Dave
1,447 views