How do you solve 6(cos^2)x+cosx=1

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
1. 0 answers
2. asked by Edward
3. 980 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. 609 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. 777 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. 1,076 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. 1,088 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. 720 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. 950 views
I have a question relating to limits that I solvedlim(x-->0) (1-cosx)/2x^2 I multiplied the numerator and denominator by
1. 1 answer
2. asked by Alex
3. 692 views
Trigonometric IdentitiesProve: (tanx + secx -1)/(tanx - secx + 1)= tanx + secx My work so far: (sinx/cosx + 1/cosx +
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2. asked by Dave
3. 1,451 views
sin2x+cosx=0 , [-180,180)= 2sinxcosx+cosx=0 = cosx(2sinx+1)=0 cosx=0 x1=cos^-1(0) x1=90 x2=360-90 x2=270 270 doesn't fit in
1. 1 answer
2. asked by Anonymous
3. 918 views