2(sinx)^2-5cosx-4=0 Solve for all solutions

2(sinx)^2-5cosx-4=0Solve for all solutions between [0,2pi]
1. 2 answers
2. asked by Anonymous
3. 776 views
Compute inverse functions to four significant digits.cos^2x=3-5cosx rewrite it as.. cos^2x + 5cosx -3=0 now you have a
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2. asked by Kate
3. 625 views
Prove the following:[1+sinx]/[1+cscx]=tanx/secx =[1+sinx]/[1+1/sinx] =[1+sinx]/[(sinx+1)/sinx] =[1+sinx]*[sinx/(sinx+1)]
1. 3 answers
2. asked by Anonymous
3. 944 views
x^3(sinx)/ 5cosx
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2. asked by jnky
3. 528 views
2(sinx)^2-5cosx-4=0
1. 1 answer
2. asked by Anonymous
3. 395 views
y = x^3 sinx / 5cosx
1. 1 answer
2. asked by jnky
3. 447 views
Hello! Can someone please check and see if I did this right? Thanks! :)Directions: Find the exact solutions of the equation in
1. 1 answer
2. asked by Maggie
3. 834 views
I need help solving for all solutions for this problem:cos 2x+ sin x= 0 I substituted cos 2x for cos^2x-sin^2x So it became
1. 1 answer
2. asked by Martha
3. 633 views
1.Solve, finding all solutions in [0, 2π).cosx sin2x + sinx cosx - sinx = 0 2.Solve, finding all solutions in [0, 2π). 3sec x
1. 1 answer
2. asked by brittney
3. 779 views
Compute inverse functions to four significant digits.cos^2x=3-5cosx cos^2x + 5cosx -3=0 now you have a quadratic, solve for cos
1. 0 answers
2. asked by Kate
3. 556 views