sin2x+cosx=0 , [-180,180) = 2sinxcosx+cosx=0

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. 913 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. 711 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. 771 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. 1,081 views
this is for proving identies and its fustrating i can't solve this one question! lolx=feta (btw the first part is supposed to be
1. 3 answers
2. asked by diana
3. 525 views
Prove:sin2x / 1 - cos2x = cotx My Attempt: LS: = 2sinxcosx / - 1 - (1 - 2sin^2x) = 2sinxcosx / - 1 + 2sin^2x = cosx / sinx - 1 =
1. 5 answers
2. asked by Anonymous
3. 1,438 views
If y=3/(sinx+cosx) , find dy/dxA. 3sinx-3cosx B. 3/(sinx+cosx)^2 C. -3/(sinx+cosx)^2 D. 3(cosx-sinx)/(sinx+cosx)^2 E.
1. 1 answer
2. asked by Anonymous
3. 663 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. 1,069 views
Verify the identity:tanx(cos2x) = sin2x - tanx Left Side = (sinx/cosx)(2cos^2 x -1) =sinx(2cos^2 x - 1)/cosx Right Side = 2sinx
1. 0 answers
2. asked by Ashley
3. 1,242 views
Is this correct?(sinx - cosx)^2 = 1 - 2sinxcosx LS = sin^2x - 2sinxcosx + cos^2x = 1 - cos^2x - 2sinxcosx + cos^2x = 1 - cos^2x
1. 1 answer
2. asked by Chewbacca
3. 625 views