Solve for [0, 360) 2sinxcosx

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. 627 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. 0 answers
2. asked by Anonymous
3. 565 views
solve algebraically, express the roots in exact form2sinxcosx-1=0 I got: 2sinxcosx=1 sin^2x=1 what do I do next? thanks in
1. 0 answers
2. asked by sh
3. 492 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. 526 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. 1,087 views
Solve identity,(1-sin2x)/cos2x = cos2x/(1+sin2x) I tried starting from the right side, RS: =(cos²x-sin²x)/(1+2sinxcosx)
1. 3 answers
2. asked by sh
3. 749 views
Prove the identity:1-cos2x/sin2x = 1+tanx/1+cotx I simplified the RS to tanx. LS:(1-2cos²x+1)/2sinxcosx (2+2cos²x)/2sinxcosx
1. 2 answers
2. asked by sh
3. 1,321 views
Solve the following equations for 0 < x < 2piA) sin^2x=2sinxcosx B) 3tanx=cosx Please help Thankyou
1. 2 answers
2. asked by Anonymous
3. 656 views
sin^2(2x)=2sinxcosx. Find all solutions to each equation in the interval [0, 2pi)So I started off changing 2sinxcosx = sin(2x),
1. 2 answers
2. asked by Anonymous
3. 1,934 views
How do I solve these?1) 2sinxcosx-cosx=0 2) cos^2(x)-0.5cosx=0 3) 6sin^2(x)-5sinx+1=0 4) tan^2(x)+tanx-12=0
1. 2 answers
2. asked by .
3. 1,812 views