solve cosx= (squareroot3)/ 2 where

hi im back..solve cosx= (squareroot3)/ 2 where 0 <(with a line underneath) x <(with a line underneath) 2 pie a) how many
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2. asked by Maria
3. 514 views
1)show that squareroot3-squareroot2=1/squareroot3+squareroot22)hence express ig(squareroot3-squareroot2) in terms of
1. 1 answer
2. asked by romaro
3. 394 views
Solve this equation fo rx in the interval 0<=x<=3603sinxtanx=8 I would do it this way: sinxtanx = 8/3 sinx(sinx/cosx)=8/3
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2. asked by Edward
3. 978 views
i am trying to solve for A and B....this question comes from a Differential equations...but this step goes back to basics...Asin
1. 2 answers
2. asked by mary
3. 605 views
solve cosx= (squareroot3)/ 2 where 0 <(with a line underneath) x <(with a line underneath) 2 piea) how many solutions are
1. 2 answers
2. asked by Maria
3. 577 views
Solve:cos(2x-180) - sin(x-90)=0 my work: cos2xcos180 + sin2xsin180= sinxcos90 - sin90cosx -cos2x - sin2x= cosx -cos^2x + sin^2x
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2. asked by Jess
3. 773 views
Simplify #3:[cosx-sin(90-x)sinx]/[cosx-cos(180-x)tanx] = [cosx-(sin90cosx-cos90sinx)sinx]/[cosx-(cos180cosx+sinx180sinx)tanx] =
1. 1 answer
2. asked by Anonymous
3. 1,072 views
Solve for [0, 360)2sinxcosx + cosx =0 2sinxcosx = -cosx 2sinx = -cosx/cosx sinx = -1/2 {210, 330) Is this correct?
1. 1 answer
2. asked by Sara
3. 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
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2. asked by elle
3. 715 views
solve each equation for 0=/<x=/<2pisin^2x + 5sinx + 6 = 0? how do i factor this and solve? 2sin^2 + sinx = 0 (2sinx - 3)(sinx
1. 7 answers
2. asked by sh
3. 947 views