solve secx/2=cosx/2 over the interval[0,2pi)

6. Solve the following equations on the interval xa) 6 cos? * + cosx - 1 = 0 b) 8 cos? x + 14 cosx - 3 c) sec? x + 2 secx - 8 =
1. 1 answer
2. 55 views
Trigonometric IdentitiesProve: (tanx + secx -1)/(tanx - secx + 1)= tanx + secx My work so far: (sinx/cosx + 1/cosx +
1. 0 answers
2. asked by Dave
3. 1,459 views
solve secx/2=cosx/2 over the interval[0,2pi) without using a calculator
1. 1 answer
2. asked by jake
3. 516 views
Write in terms of cos and sin function.cotx*secx Show work. I know cotx = cosx/sinx and secx = 1/cosx, would that just be the
1. 1 answer
2. asked by Need this asap please
3. 622 views
1/tanx-secx+ 1/tanx+secx=-2tanxso this is what I did: =tanx+secx+tanx-secx =(sinx/cosx)+ (1/cosx)+(sinx/cosx)-(1/cosx)
1. 0 answers
2. asked by olivia
3. 1,352 views
Having trouble verifying this identity...cscxtanx + secx= 2cosx This is what I've been trying (1/sinx)(sinx/cosx) + secx = 2cosx
1. 2 answers
2. asked by Alex
3. 688 views
1. (sinx/cscx)+(cosx/secx)=12. (1/sinxcosx)-(cosx/sinx)=tanx 3. (1/1+cos s)=csc^2 s-csc s cot s 4.
1. 1 answer
2. asked by Tom
3. 778 views
If secx = 8 and -pi/2 < x < 0, find the exact value of sin2xUse the identity sin 2x = 2(sinx)(cosx) if secx = 8, then cosx = 1/8
1. 0 answers
2. asked by Ashley
3. 1,184 views
I'm working this problem:∫ [1-tan^2 (x)] / [sec^2 (x)] dx ∫(1/secx)-[(sin^2x/cos^2x)/(1/cosx) ∫cosx-sinx(sinx/cosx)
1. 1 answer
2. asked by Janice
3. 710 views
help me solve this coz im stumpedsecx-secx sin^2x= cosx thanks
1. 1 answer
2. asked by Kakashi
3. 6,892 views